Entrance Corner

Hint

(b) Angular frequency \(\omega=\sqrt{\frac{k}{m}}\) Maximum velocity \(v_{\max }=A \omega\)
The angular frequency of body A is given by:
\(\omega_A=\sqrt{\frac{k_1}{m}}\)
The angular frequency of body B is given by:
\(\omega_B=\sqrt{\frac{k_2}{m}}\)
The maximum velocity of body A is given by:
\(v_{\max , A}=A \omega_A=A \sqrt{\frac{k_1}{m}}\)
The maximum velocity of body B is given by:
\(v_{\max , B}=A \omega_B=A \sqrt{\frac{k_2}{m}}\)
The ratio of the maximum velocity of \(A\) to the maximum velocity of \(B\) is:
\(
\frac{v_{\max , A}}{v_{\max , B}}=\frac{A \sqrt{\frac{k_1}{m}}}{A \sqrt{\frac{k_2}{m}}}=\sqrt{\frac{k_1}{k_2}}
\)
The ratio of the maximum velocity of \(A\) to the maximum velocity of \(B\) is \(\sqrt{\frac{k_1}{k_2}}\).

Hint

(c) Time period of a simple pendulum:
\(
T=2 \pi \sqrt{\frac{l}{g}}
\)
Here, \(T\) is the time period, \(l\) is the length of the pendulum, and \(g\) is the acceleration due to gravity.
When we take the pendulum clock to a mountaintop, the value of \(g\) (acceleration due to gravity) decrease because gravity is weaker at higher altitudes.
If \(g\) decreases, then \(\sqrt{\frac{l}{g}}\) increases, which means that \(T\) (the time period) increases.
Since the time period \(T\) increases, the pendulum clock will gain time (i.e., it will run slower compared to a clock at sea level).
Therefore, Statement A is true.
Statement R states that the value of acceleration due to gravity is lower at the mountaintop than at the plane.
This is a known fact in physics: as altitude increases, the value of \(g\) decreases.
Thus, Statement R is also true.
Therefore, Both statements are true, and Statement R is the correct explanation for Statement A.

The acceleration due to gravity (g) decreases with increasing height, following an inverse square relationship with the distance from the Earth’s center, described by the equation:
\(
g_h=g_e \times \frac{R_e{ }^2}{\left(R_p+h\right)^2}
\)
where \(g_h\) is the gravity at height \({h},{g_e}\) is gravity at the surface of the earth, and \(R_e\) is the Earth’s radius.

Hint

(d) Initial position: \(x_0\)
Initial momentum: \({p}_0\)
Angular frequency: \(\omega\)
Position in SHM: \({x}({t})={A} \cos (\omega t+{\phi})\)
Velocity in SHM: \(v(t)=-A \omega \sin (\omega t+\phi)\)
Momentum: \(p(t)=m v(t)\)

Step 1: Express initial position and momentum.
At \(t=0, x(0)=x_0=A \cos (\phi)\).
At \(t=0, p(0)=p_0=m v_0=-m A \omega \sin (\phi)\).

Step 2: Solve for amplitude \({A}\).
Square \(x_0\) and \(\frac{p_0}{m \omega}\) and add them:
\(
\begin{aligned}
& x_0^2+\left(\frac{p_0}{m \omega}\right)^2=A^2 \cos ^2(\phi)+A^2 \sin ^2(\phi)=A^2\left(\cos ^2(\phi)+\sin ^2(\phi)\right)=A^2 \\
& \text { Therefore, } A=\sqrt{x_0^2+\left(\frac{p_0}{m \omega}\right)^2}
\end{aligned}
\)

Step 3: Solve for phase \(\phi\).
\(
\begin{aligned}
& \text { Divide } \frac{p_0}{m \omega} \text { by } x_0: \\
& \frac{p_0}{m \omega x_0}=-\frac{A \sin (\phi)}{A \cos (\phi)}=-\tan (\phi)
\end{aligned}
\)
Therefore, \(\phi=\arctan \left(-\frac{p_0}{m \omega x_0}\right)\).
Determine position and momentum at any time \(\boldsymbol{1}\).
Since \(A\) and \(\phi\) are determined by \(x_0\) and \(p_0\), we can find \(x(t)=A \cos (\omega t+\phi)\) and \(p(t)=-m A \omega \sin (\omega t+\phi)\) at any time \(t\).

Knowing the initial position and momentum is sufficient to determine the position and momentum at any time \(t\) because the amplitude and phase can be expressed in terms of the initial position and momentum.

Hint

(a)

\(
x=x_0 \sin ^2\left(\frac{t}{2}\right)\left\{\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\right\}
\)
\(
x=x_0\left(\frac{1-\cos t}{2}\right)
\)
Extreme position 1: When \(\cos t=1\), \(x=0\)
and Extreme position 2:
\(\cos t=-1, x=x_0\)
Mean position is \(x=\frac{x_0}{2}\)
We know from the kinetic energy graph it is maximum at the mean position and 0 at the extreme position.

Hint

(c) Time period: \({T}=2 \mathrm{~s}\)
Amplitude: \(A=1 {~cm}\)
Time elapsed: \({t}=12.5 \mathrm{~s}\)

Step 1: Number of oscillations \(n\) is given by:
\(n=\frac{t}{T}\)
\(n=\frac{12.5 \mathrm{~s}}{2 \mathrm{~s}}\)
\(n=6.25\)

Step 2: Total distance \({D}\) is given by:
\(D=n \times 4 A\)
\(D=6.25 \times 4 \times 1 \mathrm{~cm}\)
\(D=25 \mathrm{~cm}\)

Step 3: Calculate the displacement
After 6 complete oscillations, the particle returns to its initial position.
The remaining 0.25 oscillation corresponds to \(\frac{1}{4}\) of the period.
In \(\frac{1}{4}\) of the period, the particle moves from the mean position to one extreme.
Displacement \(d\) is equal to the amplitude:
\({d}={A}\)
\(d=1 \mathrm{~cm}\)
\(
\begin{aligned}
& \frac{D}{d}=\frac{25 \mathrm{~cm}}{1 \mathrm{~cm}} \\
& \frac{D}{d}=25
\end{aligned}
\)
The ratio of total distance to displacement is 25.

Hint

(a) Side length of the cube: \(l=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Mass of the cube: \(\boldsymbol{m}=10 \mathrm{~g}=\mathbf{0 . 0 1} \mathrm{kg}\)
Acceleration due to gravity: \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\)
Density of water: \(\rho=10^3 \mathrm{~kg} / \mathrm{m}^3\)
Time period of oscillations: \(T=y \pi \times 10^{-2} \mathrm{~s}\)
The cross-sectional area of the cube (face area),
\(
A=l^2=(0.1)^2=0.01 \mathrm{~m}^2
\)
The time period of simple harmonic motion is given by \(T=2 \pi \sqrt{\frac{m}{k}}\), where \(m\) is the mass and \(k\) is the spring constant.
The buoyant force is given by \({F}_{l{B}}=l{\rho V g}\), where \(\rho\) is the density of the fluid, \({V}\) is the volume of the displaced fluid, and \(g\) is the acceleration due to gravity.

Step 1: Find the spring constant \(k\)
When the cube is pushed down by a small distance \(x\), the extra buoyant force is the restoring force.
The extra buoyant force is given by:
\(F=\rho\left(l^2 x\right) g\)
The spring constant \(k\) is defined as the restoring force per unit displacement:
\(k=\frac{F}{x}=\rho l^2 g\)
Substituting the given values:
\(k=\left(10^3 \mathrm{~kg} / \mathrm{m}^3\right)(0.1 \mathrm{~m})^2\left(10 \mathrm{~m} / \mathrm{s}^2\right)=100 \mathrm{~N} / \mathrm{m}\)

Step 2: The time period of oscillation is given by:
\(T=2 \pi \sqrt{\frac{m}{k}}\)
Substituting the values of \(m\) and \(k\) :
\(T=2 \pi \sqrt{\frac{0.01 \mathrm{~kg}}{100 \mathrm{~N} / \mathrm{m}}}=2 \pi \sqrt{10^{-4} \mathrm{~s}^2}=2 \pi \times 10^{-2} \mathrm{~s}\)

Step 3: Solve for \(y\)
Given that \(T=y \pi \times 10^{-2} \mathrm{~s}\), and we found \(T=2 \pi \times 10^{-2} \mathrm{~s}\) :
\(y \pi \times 10^{-2}=2 \pi \times 10^{-2}\)
\(y=2\)

Hint

(d) The mass of the planet is 4 times the mass of the Earth.
The radius of the planet is 2 times the radius of the Earth.
The time period of a simple pendulum is given by \(T=2 \pi \sqrt{\frac{l}{g}}\), where \(l\) is the length of the pendulum and \(g\) is the acceleration due to gravity.
The acceleration due to gravity on a planet is given by \(g=\frac{{G M}}{{R}^2}\), where \({G}\) is the gravitational constant, \({M}\) is the mass of the planet, and \(R\) is the radius of the planet.

Step 1: Calculate the acceleration due to gravity on the new planet
Let \(M_e\) and \(R_e\) be the mass and radius of the Earth, respectively.
Let \({M}_{{p}}\) and \({R}_{{p}}\) be the mass and radius of the planet, respectively.
\(
\begin{aligned}
& M_p=4 M_e \\
& R_p=2 R_e
\end{aligned}
\)
The acceleration due to gravity on Earth is \(g_e=\frac{G M_e}{R_e^2}\).
The acceleration due to gravity on the planet is \(g_p=\frac{G M_p}{R_p^2}\).
\(
g_p=\frac{G\left(4 M_e\right)}{\left(2 R_e\right)^2}=\frac{4 G M_e}{4 R_e^2}=\frac{G M_e}{R_e^2}=g_e
\)

Step 2: Calculate the time period of the pendulum on the new planet
The time period of the pendulum on Earth is \(T_e=2 \pi \sqrt{\frac{l}{g_e}}\).
The time period of the pendulum on the planet is \(T_p=2 \pi \sqrt{\frac{l}{g_p}}\).
Since \(g_p=g_e\), then \(T_p=2 \pi \sqrt{\frac{l}{g_e}}=T_e\).

Step 3: Analyze the statements
Assertion (A) is true because the time period of the pendulum remains the same on Earth and the planet.
Reason (R) is true because the mass of the pendulum remains unchanged.
However, Reason (R) does not explain Assertion (A) because the time period depends on the acceleration due to gravity, not the mass of the pendulum.

Both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Hint

(a) Time period at height \(R: T_1=4 \mathrm{~s}\)
Height 1: \({h}_1={R}\)
Height 2: \(h_2=2 R\)
\({R}\) is the radius of the Earth
Time period of a simple pendulum: \(T=2 \pi \sqrt{\frac{l}{g}}\)
Acceleration due to gravity at height \(h: g_h=\frac{G M}{(R+h)^2}\)

Step 1: Find the acceleration due to gravity at height \(R\)
The acceleration due to gravity at height \(h_1={R}\) is:
\(g_1=\frac{G M}{(R+R)^2}=\frac{G M}{4 R^2}\)

Step 2: Find the acceleration due to gravity at height \(2 R\)
The acceleration due to gravity at height \(h_2=2 R\) is:
\(
g_2=\frac{G M}{(R+2 R)^2}=\frac{G M}{9 R^2}
\)

Step 3: Find the time period \(T_1\) at height \(R\)
The time period \(T_1\) at height \(R\) is:
\(T_1=2 \pi \sqrt{\frac{l}{g_1}}=2 \pi \sqrt{\frac{4 l R^2}{G M}}\)

Step 4: Find the time period \(T_2\) at height \(2 R\)
The time period \(T_2\) at height \(2 R\) is:
\(
T_2=2 \pi \sqrt{\frac{l}{g_2}}=2 \pi \sqrt{\frac{9 l R^2}{G M}}
\)

Step 5: Find the ratio of \(T_2\) to \(T_1\)
The ratio of \(T_2\) to \(T_1\) is:
\(\frac{T_2}{T_1}=\frac{2 \pi \sqrt{\frac{\varphi R^2}{G M}}}{2 \pi \sqrt{\frac{4 R^2}{G M}}}\)
\(
\frac{T_2}{T_1}=\frac{3}{2}
\)

Step 6: Express the relationship between \(T_1\) and \(T_2\)
The relationship between \(T_1\) and \(T_2\) is:
\(2 T_2=3 T_1\)

Hint

(d) Initial total mechanical energy: \({E}\)
Mass of the oscillating particle is doubled: \({m}^{\prime}=2 m\)
Amplitude remains constant
The total mechanical energy of a system in simple harmonic motion is given by \(E=\frac{1}{2} m \omega^2 A^2\), where \(m\) is the mass, \(\omega\) is the angular frequency, and \(A\) is the amplitude.
The angular frequency \(\omega\) is related to the spring constant \(k\) and mass \(m\) by \(\omega=\sqrt{\frac{k}{m}}\)

Step 1: Express the initial total mechanical energy
The initial total mechanical energy is given by:
\(E=\frac{1}{2} m \omega^2 A^2\)
\(E=\frac{1}{2} m\left(\frac{k}{m}\right) A^2\)
\(E=\frac{1}{2} k A^2\) (Energy is independent of mass)

Step 2: Express the new total mechanical energy
When the mass is doubled, the new mass is \(m^{\prime}=2 m\).
The new angular frequency is \(\omega^{\prime}=\sqrt{\frac{k}{2 m}}\).
The new total mechanical energy \({E}^{\prime}\) is:
\(
E^{\prime}=\frac{1}{2} m^{\prime}\left(\omega^{\prime}\right)^2 A^2=\frac{1}{2}(2 m)\left(\frac{k}{2 m}\right) A^2=\frac{1}{2} k A^2
\)
Therefore, \({E}^{\prime}=E\) (reason is Energy is independent of mass)

Hint

(a) Length of pendulum: \(l=1 \mathrm{~m}\)
Mass of bob: \(m_b=1 \mathrm{~kg}\)
Mass of bullet: \({m}_H=10^{-2} \mathrm{~kg}\)
Velocity of bullet: \(v_b=2 \times 10^2 \mathrm{~m} / \mathrm{s}\)
Acceleration due to gravity: \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\)
Conservation of momentum: \(m_1 v_1+m_2 v_2=\left(m_1+m_2\right) v_f\)
Kinetic energy: \(K E=\frac{1}{2} m v^2\)
Potential energy: \({P E}={m g} {h}\)

How to Solve?
Calculate the velocity of the bob and bullet system immediately after the collision using conservation of momentum.
Calculate the kinetic energy of the system after the collision.
Equate the kinetic energy to the potential energy at the maximum height to find the height.

Step 1: Calculate the velocity of the bob and bullet system after the collision
Apply the conservation of momentum equation:
\(m_{B l} v_{b l}+m_b v_b=\left(m_{b l}+m_b\right) v_f\)
Since the bob is initially at rest, \(v_b=0\).
\(v_f=\frac{m_{b l} v_{b l}}{m_{b l}+m_b}\)
\(
=\frac{10^{-2} \mathrm{~kg} \times 2 \times 10^2 \mathrm{~m} / \mathrm{s}}{10^{-2} \mathrm{~kg}+1 \mathrm{~kg}}\approx 1.98 \mathrm{~m} / \mathrm{s}
\)

Step 2: Calculate the kinetic energy of the system after the collision
Use the kinetic energy formula:
\(K E=\frac{1}{2}\left(m_{bl}+m_b\right) v_f^2\)
\(
=\frac{1}{2}\left(10^{-2} \mathrm{~kg}+1 \mathrm{~kg}\right)(1.98 \mathrm{~m} / \mathrm{s})^2\approx 1.98 \mathrm{~J}
\)

Step 3: Calculate the height the bob rises
Equate the kinetic energy to the potential energy:
\(K E=P E\)
\(\frac{1}{2}\left(m_{b l}+m_b\right) v_f^2=\left(m_{b l}+m_b\right) g h\)

Solve for \(h\) : \(h=\frac{v_f^2}{2 g}\)
Substitute the values:
\(h=\frac{(1.98 \mathrm{~m} / \mathrm{s})^2}{2 \times 10 \mathrm{~m} / \mathrm{s}^2}\)
\(h=\frac{3.9204 \mathrm{~m}^2 / \mathrm{s}^2}{20 \mathrm{~m} / \mathrm{s}^2}\)
\(h \approx 0.196 \mathrm{~m}\)
The height to which the bob rises is approximately 0.196 m (approx 0.20 m)

Hint

(d) Length of pendulum: \(l=10 \mathrm{~m}\)
Energy dissipation: 10%
Acceleration due to gravity: \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\)
Potential energy at height \(h: P E=m g h\)
Kinetic energy: \(K E=\frac{1}{2} m v^2\)
Conservation of energy: Initial energy = Final energy + Energy dissipated

How to solve?
Apply the conservation of energy principle, accounting for energy loss due to air resistance, to find the final velocity of the bob.

Step 1: Calculate the initial potential energy of the bob.
The initial potential energy is given by:
\(
\begin{aligned}
& P E_{\text {initial }}=m g h \\
& P E_{\text {initial }}=m g l
\end{aligned}
\)
\(
P E_{\text {initial }}=m \cdot 10 \mathrm{~m} / \mathrm{s}^2 \cdot 10 \mathrm{~m}=100 \mathrm{~m} \mathrm{~J}
\)

Step 2: Calculate the energy dissipated due to air resistance.
The energy dissipated is \(10 \%\) of the initial potential energy:
\(E_{\text {dissipated }}=0.10 \cdot P E_{\text {initial }}\)
\(E_{\text {dissipated }}=0.10 \cdot 100 \mathrm{~m} \mathrm{~J}\)
\(E_{\text {dissipated }}=10 \mathrm{~m} \mathrm{~J}\)

Step 3: Calculate the final kinetic energy of the bob.
The final kinetic energy is the initial potential energy minus the energy dissipated:
\(K E_{\text {final }}=P E_{\text {initial }}-E_{\text {dissipated }}\)
\(K E_{\text {final }}=100 \mathrm{~m} \mathrm{~J}-10 \mathrm{~m} \mathrm{~J}\)
\(K E_{\text {final }}=90 \mathrm{~m} \mathrm{~J}\)

Step 4: Calculate the final velocity of the bob.
The final kinetic energy is also given by:
\(K E_{\text {final }}=\frac{1}{2} m v^2\)
Equating the two expressions for \(\boldsymbol{K} E_{\text {final }}\) :
\(\frac{1}{2} m v^2=90 m \mathrm{~J}\)
\(
v^2=\frac{2 \cdot 90 m \mathrm{~J}}{m}
\)
\(
v=6 \sqrt{5} \mathrm{~m} / \mathrm{s}
\)
The speed with which the bob arrives at the lowest point is \(6 \sqrt{5} \mathrm{~m} / \mathrm{s}\).

Hint

(c)  (A), (B), and (C) are correct because in SHM, the restoring force is directly proportional to the displacement, the acceleration and displacement are opposite in direction, and the velocity is maximum at the mean position.

Explanation:

In SHM, the restoring force \(F\) is given by Hooke’s Law, which states that \(F=-k x\), where \(k\) is the spring constant and \(x\) is the displacement from the mean position. The negative sign indicates that the force acts in the opposite direction of the displacement. Thus, the restoring force is indeed directly proportional to the displacement.

The acceleration \(a\) in SHM is given by \(a=\frac{F}{m}=\frac{-k x}{m}\). This shows that acceleration is also proportional to the displacement but in the opposite direction (due to the negative sign). Therefore, the acceleration and displacement are indeed opposite in direction.

In SHM, the velocity \(v\) is maximum at the mean position (equilibrium position) because this is where the object has the maximum kinetic energy and minimum potential energy. As the object moves away from the mean position, the velocity decreases to zero at the extreme positions. Velocity is given by \(v=\omega \sqrt{A^2-x^2}\) and it is maximum at mean position where \(x=0\). \(v_{max}=\omega A\).

At the extreme points of SHM, the displacement is maximum, which means the restoring force (and hence the acceleration) is also maximum. Therefore, the statement that acceleration is minimum at extreme points is incorrect. Acceleration is given by \(a=-\omega^2 x\).
\(
\begin{aligned}
&\text { When the particle is at the extreme position, i.e. } x=A \text {, then acceleration is maximum. }\\
&\therefore \quad a_{(\text {extreme position })}=-\omega^2 A
\end{aligned}
\)

Hint

(a) Kinetic energy in SHM: \(K E=\frac{1}{2} m \omega^2\left(A^2-x^2\right)\)
Potential energy in SHM: \(P E=\frac{1}{2} m \omega^2 x^2\) \(x\) is the displacement from the mean position.
How to solve?
Equate the kinetic and potential energies and solve for \(x\).

Step 1: Equate kinetic and potential energy
Set \(K E=P E\) :
\(\frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{1}{2} m \omega^2 x^2\)
\(
A^2-x^2=x^2
\)
\(
x^2=\frac{A^2}{2}
\)

Step 2: Solve for \({x}\)
Take the square root of both sides:
\(x=\sqrt{\frac{A^2}{2}}\)
\(x=\frac{A}{\sqrt{2}}\)

The distance from the mean position is \(\frac{A}{\sqrt{2}}\).

Hint

(d) In Simple Harmonic Motion, a particle oscillates about a mean position. The total mechanical energy ( \(E\) ) of the system remains constant and is the sum of kinetic energy \(({KE})\) and potential energy (\(U\)):
\(
E=KE+U
\)
The potential energy \(({U})\) of a particle in SHM can be expressed as:
\(
U=\frac{1}{2} k x^2
\)
where \(k\) is the spring constant and \(x\) is the displacement from the mean position.
The kinetic energy \(({KE})\) can be derived from the total energy:
\(
KE=E-U
\)
Therefore, substituting for \(({U})\) :
\(
KE=E-\frac{1}{2} k x^2
\)
Since \(KE=E-\frac{1}{2} k x^2\), we can see that as \(x\) increases (moving away from the mean position), the potential energy increases, and consequently, the kinetic energy decreases.
The graph of kinetic energy ( \(KE\) ) versus displacement ( \(x\) ) will be a downward opening parabola.
The graph that represents the difference between total energy and potential energy (which is the kinetic energy) will be a downward parabola, reaching its maximum at the mean position (where \(x=0\) ) and decreasing to zero when the particle reaches the maximum displacement (amplitude).

Hint

(d) Given, displacement \(x\) is half the amplitude \(A: x=\frac{A}{2}\).
Potential energy in SHM: \(U=\frac{1}{2} k x^2\).
Kinetic energy in SHM: \(K=\frac{1}{2} k\left(A^2-x^2\right)\).
How to solve?
Calculate the potential and kinetic energies using the given displacement and then find their ratio.

Step 1: Calculate the potential energy \({U}\).
Substitute \(x=\frac{A}{2}\) into the potential energy formula:
\(U=\frac{1}{2} k\left(\frac{A}{2}\right)^2\)
\(U=\frac{1}{8} k A^2\)

Step 2: Calculate the kinetic energy \(K\).
Substitute \(x=\frac{{A}}{2}\) into the kinetic energy formula:
\(K=\frac{1}{2} k\left(A^2-\left(\frac{A}{2}\right)^2\right)\)
\(
K=\frac{3}{8} k A^2
\)

Step 3: Find the ratio of potential energy to kinetic energy.
Divide the potential energy by the kinetic energy:
\(\frac{U}{K}=\frac{\frac{1}{8} k A^2}{\frac{3}{8} k A^2}\)
\(\frac{U}{K}=\frac{1}{3}\)

The ratio of potential energy to kinetic energy is \(1: 3\).

Hint

(c) \(K E=\frac{1}{2} k\left(A^2-x^2\right)\)
This shows that KE depends on \(A^2-x^2\).
As the particle moves from the mean position \((x=0)\) to the extreme position ( \({x}={A}\) ), the value of \(x^2\) increases. Consequently, \(A^2-x^2\) decreases, which means that KE decreases as \(x\) increases from \(O\) to \(A\).
The relationship \(K E \propto A^2-x^2\) indicates that the graph of KE vs. \(x\) will be a downward-opening parabola.
At the mean position \((x=0)\), KE is maximum, and at the extreme position ( \(\mathrm{x}=\mathrm{A}\) ), KE is zero.
Therefore the variation of kinetic energy with displacement from the mean position to the extreme position is parabolic and decreases as the displacement increases.

Hint

(d) Amplitude of SHM: \(A\)
Position at \(t=0: x=\frac{A}{2}\)

Direction of motion at \(t=0\) : positive x-axis
General equation for SHM: \({x}=A \sin (\omega t+\delta)\)

How to solve?
Substitute the given values into the SHM equation and solve for the phase constant \(\delta\).

Step 1: Substitute \(t=0\) and \(x=\frac{{A}}{2}\) into the SHM equation
\(
\begin{aligned}
& x=A \sin (\omega t+\delta) \\
& \frac{A}{2}=A \sin (\omega \cdot 0+\delta) \\
& \frac{1}{2}=\sin (\delta)
\end{aligned}
\)

Step 2: Find the possible values of \(\delta\)
\(
\delta=\frac{\pi}{6} \text { or } \delta=\frac{5 \pi}{6}
\)
Therefore, \(\delta=\frac{\pi}{6}\)
The phase constant is \(\frac{\pi}{6}\).

Hint

(c) After time \(t\), the angular displacement will be
\(
\theta=\omega t
\)
Total angular displacement from \(x\)-axis.
\(
\theta_{\text {total }}=\omega t+\frac{\pi}{6} \quad\left(\because 30^{\circ}=\frac{\pi}{6}\right)
\)
Now, OP has two component
The horizontal component will be the projection along \(x\)-axis
\(
=r \cos \left(\theta_{\text {total }}\right)=r \cos \left(\omega t+\frac{\pi}{6}\right)
\)

Hint

(a) Both the springs are in parallel.
\(
\begin{aligned}
& K_{e q}=K_1+K_2 \\
& \text { Time period of oscillation, } T=2 \pi \sqrt{\frac{m}{K_{e q}}}=2 \pi \sqrt{\frac{m}{K_1+K_2}}
\end{aligned}
\)

Hint

(a) The time period \(T\) of a simple pendulum is given by the formula:
\(
T=2 \pi \sqrt{\frac{L}{g}}
\)
where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.
\(
T^2=\frac{4 \pi^2}{g} L
\)
From the equation \(T^2=\frac{4 \pi^2}{g} L\), we can see that \(T^2\) is directly proportional to \(L\).
This means that if we plot \(T^2\) on the \(y\)-axis and \(L\) on the x-axis, the relationship will be linear.

Since \(T^2\) is directly proportional to \(L\), the graph will be a straight line that passes through the origin \((0,0)\).
This indicates that as the length \(L\) increases, the square of the time period \(T^2\) also increases linearly.

Hint

(c) Maximum potential energy: \({U}_{\max }=25 \mathrm{~J}\)
Displacement: \({x}=\frac{{A}}{2}\) where \({A}\) is the amplitude
Total energy in SHM: \({E}={U}+{K}\)
Maximum potential energy: \(U_{\max }=\frac{1}{2} k A^2\)
Potential energy at displacement \(\mathrm{x}: U=\frac{1}{2} k x^2\)
How to solve?
Calculate the potential energy at the given displacement and subtract it from the total energy to find the kinetic energy.

Step 1: Calculate the potential energy at \(x=\frac{A}{2}\)
The potential energy at displacement \(x\) is given by \(U=\frac{1}{2} k x^2\).
Substituting \(x=\frac{A}{2}\), we get \(U=\frac{1}{2} k\left(\frac{A}{2}\right)^2\).
Simplifying, \(U=\frac{1}{8} k A^2\).
Since \(U_{\text {max }}=\frac{1}{2} k A^2=25 \mathrm{~J}\), then \(k A^2=2 \cdot 25 \mathrm{~J}=50 \mathrm{~J}\).
Substituting \(k A^2\) into the equation for \(U\), we get \(U=\frac{1}{8}(50 \mathrm{~J})=\frac{25}{4} \mathrm{~J}=6.25 \mathrm{~J}\).

Step 2: Calculate the kinetic energy
The total energy \(E\) is equal to the maximum potential energy, so \(E={U}_{\max }=25 \mathrm{~J}\)
The kinetic energy \({K}\) is given by \({K}={E}-{U}\).
Substituting the values, \(K=25 \mathrm{~J}-6.25 \mathrm{~J}=18.75 \mathrm{~J}\).
The kinetic energy of the block at \(x=\frac{A}{2}\) is 18.75 J.

Hint

(a)

\(
\begin{aligned}
& \omega=\sqrt{\frac{K}{m}} \Rightarrow \omega \propto \frac{1}{\sqrt{m}} \\
& \frac{\omega_2}{\omega_1}=\sqrt{\frac{m_1}{m_2}}=\sqrt{\frac{1}{2}}
\end{aligned}
\)

Hint

(a) The particle is in simple harmonic motion between – \({A}\) and \({A}\).
Time taken to go from 0 to \(\frac{A}{2}\) is 2 s.

The displacement equation for SHM is \({x}={A} \sin (\omega t)\).
The time period \(T\) is related to angular frequency \(\omega\) by \(T=\frac{2 \pi}{\omega}\).
How to solve?
Use the displacement equation to find the time taken to reach \(\frac{A}{2}\) and \(A\), then calculate the difference.

Step 1: Find the time \(t_1\) to reach \(\frac{A}{2}\)
Set \(x=\frac{{A}}{2}\) in the displacement equation:
\(\frac{A}{2}=A \sin \left(\omega t_1\right)\)
Simplify:
\(\frac{1}{2}=\sin \left(\omega t_1\right)\)
Solve for \(\omega t_1\) :
\(\omega t_1=\frac{\pi}{6}\)
Given \(t_1=2 \mathrm{~s}\), find \(\omega\) :
\(\omega=\frac{\pi}{6 t_1}=\frac{\pi}{6(2)}=\frac{\pi}{12} \mathrm{rad} / \mathrm{s}\)

Step 2: Find the time \(t_2\) to reach \({A}\)
Set \({x}={A}\) in the displacement equation:
\(A=A \sin \left(\omega t_2\right)\)
Simplify:
\(1=\sin \left(\omega t_2\right)\)
Solve for \(\omega t_2\) :
\(\omega t_2=\frac{\pi}{2}\)
Find \(t_2\) :
\(t_2=\frac{\pi}{2 \omega}=\frac{\pi}{2\left(\frac{\pi}{12}\right)}=6 \mathrm{~s}\)

Step 3: Calculate the time difference
Find the time taken to go from \(\frac{A}{2}\) to \(A\) :
\(\Delta t=t_2-t_1=6-2=4 \mathrm{~s}\)

The time taken by the particle to go from \(\frac{A}{2}\) to \(A\) is 4 s .

Hint

(c) Time period on Earth’s surface: \({T}\)
Time period at height \({R}: {x} {T}\)
Height above Earth’s surface: \({R}\) (equal to Earth’s radius)

Time period of a simple pendulum: \(T=2 \pi \sqrt{\frac{l}{g}}\)
Acceleration due to gravity at height \(h: g_h=\frac{g}{\left(1+\frac{h}{R}\right)^2}\)

How to solve?
Calculate the acceleration due to gravity at height \(R\), then find the new time period and compare it to the original time period.

Step 1: Calculate the acceleration due to gravity at height \({R}\).
Substitute \(h={R}\) into the formula for \(g_h\) :
\(g_R=\frac{g}{\left(1+\frac{R}{R}\right)^2}\)
\(g_R=\frac{g}{(1+1)^2}\)
\(g_R=\frac{g}{4}\)

Step 2: Calculate the time period at height \(R\).
Use the formula for the time period of a simple pendulum with \(g_{R}\) :
\(T_R=2 \pi \sqrt{\frac{l}{g_R}}=2 \pi \sqrt{\frac{l}{\frac{g}{4}}}\)
\(
\begin{aligned}
& T_R=2 \cdot 2 \pi \sqrt{\frac{l}{g}} \\
& T_R=2 T
\end{aligned}
\)

Find the value of \(x\).
Compare \(T_R\) with \(x T\) :
\(x T=2 T\)
\(x=2\)

Hint

(a) 

Length of the pendulum: \(L\)
Inclination of the plane: \({\alpha}\)
Acceleration due to gravity: \(g\)
The effective acceleration due to gravity acting on the pendulum is \({g} \cos \alpha\).

The net force on bob \(=m g \cos (\alpha)\) \(g_{e f f}=g \cos (\alpha)\).

How to solve?
Calculate the time period using the formula for a simple pendulum, replacing \(g\) with the effective acceleration \({g} \cos \alpha\).

Step 1: Calculate the effective acceleration
The effective acceleration is the component of gravity acting perpendicular to the inclined plane.
\(
a_{e f f}=g \cos \alpha
\)

Step 2: Calculate the time period
The time period of a simple pendulum is given by:
\(
T=2 \pi \sqrt{\frac{L}{a_{e f f}}}
\)
Substituting \(a_{e f f}=g \cos \alpha\) :
\(
T=2 \pi \sqrt{\frac{L}{g \cos \alpha}}
\)
The time period of oscillation of the pendulum is \(2 \pi \sqrt{\frac{L}{g \cos \alpha}}\).

Hint

(c) Time period of Clock – 1 on Earth: \({T}_1=4 \mathrm{~s}\)
Time period of Clock – 2 in space station: \({T}_{{2}}=6 \mathrm{~s}\)
Radius of Earth: \(R=6.4 \times 10^6 \mathrm{~m}\)

Time period of a simple pendulum is given by \(T=2 \pi \sqrt{\frac{l}{g}}\).
Acceleration due to gravity at height \(h\) is given by \(g_h=\frac{g R^2}{(R+h)^2}\).

How to solve?
Relate the time periods to the gravitational acceleration, then solve for \(h\).

Step 1: Relate the time periods to the gravitational accelerations
The time period of a simple pendulum is given by:
\(T_1=2 \pi \sqrt{\frac{l}{g}}\)
\(T_2=2 \pi \sqrt{\frac{l}{g_h}}\)
Divide the two equations:
\(\frac{T_2}{T_1}=\sqrt{\frac{g}{g_h}}\)

Step 2: Substitute the expression for \(g_h\)
Substitute \(g_h=\frac{g R^2}{(R+h)^2}\) into the equation:
\(\frac{T_2}{T_1}=\sqrt{\frac{g}{\frac{g R^2}{(R+h)^2}}}\)
\(\frac{T_2}{T_1}=\frac{R+h}{R}\)

Step 3: Solve for \(h\)
Rearrange the equation to solve for \(h\) :
\(R+h=R \frac{T_2}{T_1}\)
\(h=R\left(\frac{T_2}{T_1}-1\right)\)

Step 4: Substitute the given values
Substitute \(T_1=4 \mathrm{~s}, T_2=6 \mathrm{~s}\), and \(R=6.4 \times 10^6 \mathrm{~m}\) :
\(h=\left(6.4 \times 10^6 \mathrm{~m}\right)\left(\frac{6}{4}-1\right)\)
\(h=\left(6.4 \times 10^6 \mathrm{~m}\right)\left(\frac{1}{2}\right)\)
\(h=3.2 \times 10^6 \mathrm{~m}\)

The height of the space station above the Earth’s surface is \(3.2 \times 10^6 \mathrm{~m}\).

Hint

(b) For simple harmonic motion, \({x}={A} \sin \omega {t}\)
\({v}=\frac{d x}{d t}={A} \omega \cos \omega {t}\)
\(
\frac{x}{A}=\sin \omega t \dots(i)
\)
\(
\frac{{v}}{A \omega}=\cos \omega t \dots(ii)
\)
From (i) and (ii) we have,
\(
\begin{aligned}
& \left(\frac{x}{A}\right)^2+\left(\frac{{v}}{A \omega}\right)^2=\sin ^2 \omega t+\cos ^2 \omega t \\
& \left(\frac{x}{A}\right)^2+\left(\frac{{v}}{A \omega}\right)^2=1
\end{aligned}
\)
The nature of the graph will be elliptical.

Hint

(a) For System (A):
The system consists of mass \(m\) with an additional mass 2 m fixed on top, and it is attached to two springs with spring constant k on both sides.
The effective mass of the system is the sum of the masses:
\(
m_{\mathrm{eff}}=m+2 m=3 m
\)
Since the two springs are attached in parallel (both act simultaneously on the mass m\()\), the effective spring constant is: \(\left(k_{{eff}}=k+k=2 k\right)\).
The time period ( \({T}_1\) ) for this system is: \(T_1=2 \pi \sqrt{\frac{3 m}{2 k}}\)

For System (B):
The system consists of mass \(m\) attached to two springs with spring constants k and 2 k on either side.
Since the two springs are attached in parallel, the effective spring constant is: \(k_{\text {eff }}={k}+2 k=3 k\)
The effective mass remains \(m\).
The time period \(T_2\) for this system is: \(T_2=2 \pi \sqrt{\frac{m}{3 k}}\)
\(
\Rightarrow \frac{T_1}{T_2}=\frac{3}{\sqrt{2}}
\)

Hint

(c) During the time period of SHM,
\(
T=2 \pi \sqrt{\frac{l}{g}} \text { and, } \omega=\frac{2 \pi}{T}
\)
Hence, \(\omega=\frac{2 \pi}{2 \pi} \sqrt{\frac{g}{l}}\)
\(
\begin{aligned}
& \omega=\sqrt{\frac{g}{l}} \\
& l=\frac{g}{\omega^2}
\end{aligned}
\)
By comparing with the above equation, \({y}={A} \sin (\pi {t}+\phi)\)we get;
\(
\omega=\pi
\)
Hence, \(l=\frac{g}{\pi^2} \approx 99.4 \mathrm{~cm}\)

Hint

(a)

\(
\begin{aligned}
&\begin{aligned}
& x=4 \sin \left(\frac{\pi}{2}-\omega t\right) \\
& =4 \cos (\omega t) \\
& y=4 \sin (\omega t) \\
& \Rightarrow x^2+y^2=4^2
\end{aligned}\\
&\Rightarrow \text { The particle is moving in a circular motion with radius of } 4 \mathrm{~m} \text {. }
\end{aligned}
\)

Hint

(b) At \({t}=1 \mathrm{sec}\)
\(
\begin{aligned}
& x=\sin \pi\left(1+\frac{1}{3}\right) \\
& =\sin \frac{4 \pi}{3} \\
& =-\frac{\sqrt{3}}{2}
\end{aligned}
\)
So, \({V}=\omega \sqrt{{A}^2-x^2}\)
\(
\begin{aligned}
& =\pi \sqrt{1-\frac{3}{4}} \\
& =\frac{\pi}{2} \mathrm{~m} / \mathrm{s} \\
& =157 \mathrm{~cm} / \mathrm{s}
\end{aligned}
\)

Hint

(d) Time \(t=3\) seconds.
Displacement \(x=\frac{A}{2}\), where \(A\) is the amplitude.
The displacement equation for SHM starting from the mean position is \(x=A \sin (\omega t)\).
Angular frequency \(\omega=\frac{2 \pi}{T}\), where \(T\) is the time period.

How to solve
Substitute the given values into the displacement equation and solve for the time period \(T\).

Step 1: Substitute the given values into the displacement equation.
\(
\begin{aligned}
& x=A \sin (\omega t) \\
& \frac{A}{2}=A \sin (\omega \cdot 3) \\
& \frac{1}{2}=\sin (3 \omega)
\end{aligned}
\)

Step 2: Solve for \(\omega\).
\(
\begin{aligned}
& 3 \omega=\sin ^{-1}\left(\frac{1}{2}\right) \\
& 3 \omega=\frac{\pi}{6} \\
& \omega=\frac{\pi}{18}
\end{aligned}
\)

Step 3: Use the relationship between \(\omega\) and \(T\) to find \(T\).
\(
\begin{aligned}
& \omega=\frac{2 \pi}{T} \\
& \frac{\pi}{18}=\frac{2 \pi}{T} \\
& T=\frac{2 \pi}{\frac{\pi}{18}} \\
& T=2 \pi \cdot \frac{18}{\pi} \\
& T=36
\end{aligned}
\)
The time period of the harmonic motion is 36 seconds.

Hint

(c) Time period of pendulum, \(T_P=2 \pi \sqrt{\frac{\text { string length }}{\text { downward acceleration }}}=2 \pi \sqrt{\frac{\mathrm{~L}}{a_y}}\)
If the lift is stationary, \(\mathrm{a}_{\mathrm{y}}=\mathrm{g}\)
Time period is \(T=2 \pi \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}}\)
If the lift is accelerating with \(\frac{g}{6}\)
\(
\text { Force } \Rightarrow \mathrm{mg}_{\text {eff }}=\mathrm{mg}+\frac{\mathrm{mg}}{6}
\)
\(
g_{e f f}=\frac{7 g}{6}
\)
\(
\mathrm{T}=2 \pi \sqrt{\frac{L}{g_{\mathrm{eff}}}}=2 \pi \sqrt{\frac{L \times 6}{7 \mathrm{~g}}}=\sqrt{\frac{6}{7}} T
\)

Hint

(b) In \({SHM}, {v}_{\max }=\omega \mathrm{A}\)
\(
\omega \mathrm{A}=\text { constant }
\)
\(
\begin{aligned}
& \omega_1 A_1=\omega_2 A_2 \\
& \Rightarrow \frac{A_1}{A_2}=\frac{\omega_2}{\omega_1} \\
& =\sqrt{\frac{k_2}{m_2}} \times \sqrt{\frac{m_1}{k_1}}=\sqrt{\frac{9 k}{100} \times \frac{50}{2 k}}=\frac{3}{2}
\end{aligned}
\)

Hint

(b) Mass of the block: \(m=5 \mathrm{~kg}\)
Maximum potential energy: \(U_{\max }=10 \mathrm{~J}\)
Amplitude of oscillation: \({A}=2 \mathrm{~m}\)
Length of the pendulum: \(L=4 \mathrm{~m}\)
The period of the spring-mass system and the pendulum are equal.
Potential energy of a spring: \(U=\frac{1}{2} k x^2\)
Angular frequency of a spring-mass system: \(\omega=\sqrt{\frac{k}{m}}\)
Period of a spring-mass system: \(T=2 \pi \sqrt{\frac{m}{k}}\)
Period of a simple pendulum: \(T=2 \pi \sqrt{\frac{L}{g}}\)

How to solve?
Find the spring constant \(k\) using the potential energy curve.
Calculate the period of the spring-mass system.
Equate the period of the spring-mass system to the period of the pendulum.
Solve for the acceleration due to gravity \(g\).

Step 1: Find the spring constant \(k\)
The maximum potential energy is given by \(U_{\max }=\frac{1}{2} k A^2\).
Substitute the given values: \(10=\frac{1}{2} k(2)^2\).
Solve for \(k: k=\frac{2 \cdot 10}{4}=5 \frac{\mathrm{~N}}{\mathrm{~m}}\).

Step 2: Calculate the period of the spring-mass system
The period of the spring-mass system is \(T=2 \pi \sqrt{\frac{m}{k}}\).
Substitute the values: \(T=2 \pi \sqrt{\frac{5}{5}}=2 \pi \mathrm{~s}\).

Step 3: Equate the periods
The period of the pendulum is \(T=2 \pi \sqrt{\frac{L}{g}}\).
Equate the periods: \(2 \pi=2 \pi \sqrt{\frac{4}{g}}\).

Step 4: Solve for \(g\)
Divide both sides by \(2 \pi: 1=\sqrt{\frac{4}{g}}\).
Square both sides: \(1=\frac{4}{g}\).
Solve for \(g: g=4 \frac{\mathrm{~m}}{\mathrm{~s}^2}\).
The acceleration due to gravity on the planet is \(4 \frac{\mathrm{~m}}{\mathrm{~s}^2}\).

Hint

(d) Initial time period: \({T}\)
Liquid density: \(\frac{1}{4}\) of bob’s density
Length increase: \(\frac{1}{3}\) of original length

Time period of a simple pendulum: \(T=2 \pi \sqrt{\frac{l}{g}}\)
Buoyant force: \({F}_{{b}}=V_{\rho_{\text {liquid }}}g\)
Effective weight in liquid: \({W}^{\prime}={m g}-{F}_{{b}}\)

How to solve
Calculate the new length and effective gravity, then find the new time period.

Step 1: Calculate the new length
The new length \(l^{\prime}\) is the original length \(l\) plus \(\frac{1}{3}\) of \(l\) :
\(l^{\prime}=l+\frac{1}{3} l\)
\(l^{\prime}=\frac{4}{3} l\)

Step 2: Calculate the buoyant force
The buoyant force \(F_b\) is given by:
\(F_b=V \rho_{\text {liquid }}g\)
The volume \(V\) of the bob is:
\({V}=\frac{{m}}{\rho_{b o b}}\)
Since \(\rho_{\text {liquid }}=\frac{1}{4} \rho_{\text {bob }}\), the buoyant force is:
\(F_b=\frac{m}{\rho_{\text {bob }}} \cdot \frac{1}{4} \rho_{\text {bob }} \cdot g\)
\(F_b=\frac{1}{4} m g\)

Step 3: Calculate the effective weight and effective gravity
The effective weight \(W^{\prime}\) is:
\(W^{\prime}=m g-F_b\)
\(W^{\prime}=m g-\frac{1}{4} m g\)
\(W^{\prime}=\frac{3}{4} m g\)
The effective gravity \(g^{\prime}\) is:
\(g^{\prime}=\frac{W^{\prime}}{m}\)
\(g^{\prime}=\frac{\frac{3}{4} m g}{m}\)
\(g^{\prime}=\frac{3}{4} g\)

Step 4: Calculate the new time period
The new time period \({T}^{\prime}\) is:
\(T^{\prime}=2 \pi \sqrt{\frac{l^{\prime}}{g^{\prime}}}\)
\(T^{\prime}=2 \pi \sqrt{\frac{\frac{4}{3} l}{\frac{3}{4} g}}\)
\(T^{\prime}=2 \pi \sqrt{\frac{16 l}{9 g}}\)
\({T}^{\prime}=\frac{4}{3} \cdot 2 \pi \sqrt{\frac{l}{g}}\)
\(T^{\prime}=\frac{4}{3} T\)

The new time period of the simple harmonic oscillations is \(\frac{4}{3} T\).

Hint

(a) In SHM the total mechanical energy (E) is constant and is the sum of kinetic and potential energy at any point in time.
At different positions in S.H.M., the potential energy and kinetic energy can vary. They are equal at specific instances (like at the mean position), but not always.
The average kinetic and potential energies over a random time interval can differ. They are equal only over a complete cycle (one time period).
The total mechanical energy (E) is constant and is given by \(E=K . E+P\). Thus, at any point in time, the sum remains constant.

Note: In S.H.M. total mechanical energy remains constant and also \(==\frac{1}{4} K A^2\) (for one time period)

Hint

(d) From the waveform given which is a sine curve we can write
\(
x = A \sin \omega t
\)
We know Potential energy in SHM is given by
\(
U=\frac{1}{2}kx^2
\)
\(
U= \frac{1}{2}kA^2 \sin ^2 \omega t
\)
Potential energy is maximum at maximum distance from the mean position. Option (d) satisfies this.
Note: As there is \(x^2\) in PE negative half of the sine curve will become positive.

Hint

(a)

\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{m}{k}} \\
& 0.2=2 \pi \sqrt{\frac{0.5}{k}} \\
& \mathrm{k}=50 \pi^2 \\
& \approx 500 \\
& \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\phi) \\
& =5 \mathrm{~cm} \sin \left(\frac{\omega T}{4}+0\right) \\
& =5 \mathrm{~cm} \sin \left(\frac{\pi}{2}\right) \\
& =5 \mathrm{~cm} \\
& P E=\frac{1}{2} k x^2 \\
& =\frac{1}{2}(500)\left(\frac{5}{100}\right)^2 \\
& =0.6255
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& \text { Total Energy} E=\frac{1}{2} K a^2 \\
& \frac{3 E}{4}=\frac{1}{2} K\left(a^2-y^2\right) \\
& \Rightarrow \frac{3}{4} \times \frac{1}{2} K a^2=\frac{1}{2} K\left(a^2-y^2\right) \\
& \Rightarrow y^2=a^2-\frac{3 a^2}{4} \\
& \Rightarrow y=\frac{a}{2}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\text { The total mechanical energy } E \text { in SHM is constant and is given by: }\\
&E=\frac{1}{2} k A^2=\frac{1}{2} m \omega^2 A^2
\end{aligned}
\)
Kinetic Energy
\(
\begin{aligned}
& K=\frac{1}{2} m \omega^2\left(A^2-x^2\right) \\
& =\frac{1}{2} m \omega^2\left(A^2-\frac{A^2}{4}\right) \\
& =\frac{1}{2} m \omega^2\left(\frac{3 A^2}{4}\right) \\
& K=\frac{3}{4}\left(\frac{1}{2} m \omega^2 A^2\right)=\frac{3}{4} E
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&T_0=2 \pi \sqrt{\frac{l}{g}}\\
&\text { New time period } T=2 \pi \sqrt{\frac{l / 16}{g}}\\
&\begin{aligned}
& =\frac{2 \pi}{4} \sqrt{\frac{l}{g}} \\
& T=\frac{T_0}{4}
\end{aligned}
\end{aligned}
\)

Hint

(d) The velocity of a particle in SHM is given by \(v=\omega \sqrt{A^2-x^2}\), where \(\omega\) is the angular frequency, \(A\) is the amplitude, and \(x\) is the displacement from the mean position.
The time period \(T\) is related to the angular frequency \(\omega\) by \(T=\frac{2 \pi}{\omega}\).
How to solve?
Use the given velocities and positions to form two equations, eliminate the amplitude, and solve for the angular frequency, then calculate the time period.
Write the velocity equations for the two positions.
\(
\begin{aligned}
& v_1=\omega \sqrt{A^2-x_1^2} \\
& v_2=\omega \sqrt{A^2-x_2^2}
\end{aligned}
\)
\(
\begin{aligned}
& v_1^2=\omega^2\left(A^2-x_1^2\right) \\
& v_2^2=\omega^2\left(A^2-x_2^2\right)
\end{aligned}
\)
Rearrange the equations to isolate \({A}^2\).
\(
\begin{aligned}
& A^2=\frac{v_1^2}{\omega^2}+x_1^2 \\
& A^2=\frac{v_2^2}{\omega^2}+x_2^2
\end{aligned}
\)
Set the two expressions for \(A^2\) equal to each other.
\(
\frac{v_1^2}{\omega^2}+x_1^2=\frac{v_2^2}{\omega^2}+x_2^2
\)
\(
\omega^2=\frac{v_1^2-v_2^2}{x_2^2-x_1^2}
\)
\(
\omega=\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}}
\)
\(
\begin{aligned}
& T=\frac{2 \pi}{\omega} \\
& T=2 \pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}
\end{aligned}
\)
The time period of the oscillation is \(2 \pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}\).

Hint

(d) Period of the motion: \(\boldsymbol{T}=\frac{\pi}{\omega}\)

General equation for simple harmonic motion: \(x(t)=A \cos (\Omega t+\phi)\), where \(A\) is the amplitude, \(\Omega\) is the angular frequency, and \(\phi\) is the phase constant.
Relationship between period and angular frequency: \(T=\frac{2 \pi}{\Omega}\).
How to solve?
Find the angular frequency \(\Omega\) using the given period \(T\), and then compare the given options with the general form of simple harmonic motion.
Find the angular frequency \(\Omega\).
Use the formula \(T=\frac{2 \pi}{\Omega}\).
Substitute the given period \(T=\frac{\pi}{\omega}\) :
\(\frac{\pi}{\omega}=\frac{2 \pi}{\Omega}\)
Solve for \(\Omega\) :
\(\Omega=\frac{2 \pi \omega}{\pi}\)
\(\Omega=2 \omega\)

Option \(\mathrm{a}: \cos (\omega t)+\cos (2 \omega t)+\cos (3 \omega t)\) is a sum of cosine functions with different frequencies, not a single harmonic motion.
Option \(\mathrm{b}: \sin ^2(\omega t)=\frac{1}{2}-\frac{1}{2} \cos (2 \omega t)\) is a single harmonic motion with angular frequency \(2 \omega\), but the period is \(\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}\).
Option c: \(\sin (\omega t)+\cos (\omega t)=\sqrt{2} \sin \left(\omega t+\frac{\pi}{4}\right)\) is a single harmonic motion with angular frequency \(\omega\), but the period is \(\frac{2 \pi}{\omega} \neq \frac{\pi}{\omega}\).
Option d: \(3 \cos \left(\frac{\pi}{4}-2 \omega t\right)=3 \cos \left(2 \omega t-\frac{\pi}{4}\right)\) is a single harmonic motion with angular frequency \(2 \omega\), and the period is \(\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}\).

The function representing simple harmonic motion with a period of \(\frac{\pi}{\omega}\) is \(3 \cos \left(\frac{\pi}{4}-2 \omega t\right)\).

Hint

(b) 

\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{l}{g}} \\
& T^2=2 \pi\left(\frac{l}{g}\right) \\
& g=2 \pi \frac{l}{T^2} \\
& \text { The percentage error in } g \text { can be calculated using the formula: } \\
& \frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2 \Delta T}{T} \\
& \frac{\Delta g}{g}=\frac{1 \times 10^{-3}}{1 \times 10^{-2}}+\frac{2 \times 1}{100} \\
& \frac{\Delta g}{g}=0.02+0.01=0.03 \\
& 100 \times \frac{\Delta g}{g}=0.03 \times 100=3 \% \\
& \frac{\Delta g}{g} \times 100=3 \%
\end{aligned}
\)

Hint

(d)

\(
v_{\max }=A \omega
\)
Given, \(\omega_1 A_1=\omega_2 A_2\)
We know that \(\omega=\sqrt{\frac{K}{m}}\)
\(
\begin{aligned}
& \therefore \sqrt{\frac{k_1}{m}} A_1=\sqrt{\frac{k_2}{m}} A_2 \\
& \Rightarrow \frac{A_1}{A_2}=\sqrt{\frac{k_2}{k_1}}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& \mathrm{KE}=\mathrm{PE} \\
& \frac{1}{2} k\left(A^2-x^2\right)=\frac{1}{2} K x^2 \\
& \Rightarrow A^2-x^2=x^2 \\
& \Rightarrow 2 X^2=A^2 \\
& \Rightarrow x^2=\frac{A^2}{\sqrt{2}} \\
& \Rightarrow x= \pm \frac{A}{\sqrt{2}}
\end{aligned}
\)

Hint

When lift is stationary
\(
T=2 \pi \sqrt{\frac{L}{g}}
\)
A pseudo force will act downwards when lift is moving upwards.
\(
\therefore g_{e f f}=g+\frac{g}{2}=\frac{3 g}{2}
\)
\(\therefore\) New time period
\(
\begin{aligned}
& T^{\prime}=2 \pi \sqrt{\frac{L}{g_{e f f}}} \\
& T^{\prime}=2 \pi \sqrt{\frac{2 L}{3 g}} \\
& \therefore T^{\prime}=\sqrt{\frac{2}{3}} T
\end{aligned}
\)

Hint

(c) For a body performing SHM, relation between velocity and displacement
\(
v=\omega \sqrt{A^2-x^2}
\)
now, square both side
\(
\begin{aligned}
& v^2=w^2\left(A^2-x^2\right) \\
& \Rightarrow v^2=w^2 A^2-\omega^2 x^2 \\
& v^2+\omega^2 x^2=\omega^2 A^2
\end{aligned}
\)
divide whole equation by \(\omega^2 A^2\)
\(
\begin{aligned}
& \frac{v^2}{\omega^2 A^2}+\frac{\omega^2 x^2}{\omega^2 A^2}=\frac{\omega^2 x^2}{\omega^2 A^2} \\
& \frac{v^2}{(\omega A)^2}+\frac{x^2}{(A)^2}=1
\end{aligned}
\)
above equation is similar as standard equation of ellipses, so graph between velocity and displacement will be ellipses.

Hint

(b) A second’s pendulum has a time period of 2 seconds, not 1 second. Therefore, Statement I is false.
The time to move between extreme positions is half the time period.
For a second’s pendulum, this is \(\frac{2}{2}=1\) second.
Therefore, Statement II is true.

Statement I is false and Statement II is true.

Note: We know the time period of the second’s pendulum is 2 s . The time taken by particle performing SHM between two extreme positions is \({T} / 2\). For, \({T}=2 \mathrm{~s}, {~T} / 2=1 \mathrm{~s}\)

Hint

(a) 

Given, the tunnel is dug along a chord of the Earth.
The perpendicular distance from the Earth’s center to the tunnel is \(\frac{R}{2}\), where \(R\) is the radius of the Earth.
The wall of the tunnel is frictionless.
The gravitational force inside the Earth is proportional to the distance from the center.
The time period of a simple harmonic motion is given by \(T=2 \pi \sqrt{\frac{m}{k}}\), where \(m\) is the mass and \(k\) is the spring constant.
The acceleration due to gravity at the Earth’s surface is \(g=\frac{G M}{R^2}\), where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.

How to solve?
Find the gravitational force acting on the particle inside the tunnel.
Determine the component of the gravitational force along the tunnel.
Show that the motion is simple harmonic and find the effective spring constant.
Calculate the time period of the simple harmonic motion.

Step 1: Find the gravitational force at a distance \(r\) from the center of the Earth
The gravitational force inside the Earth is proportional to the distance from the center:
\(F=-\frac{G M m}{R^3} r\)
where \(m\) is the mass of the particle and \(r\) is the distance from the center of the Earth.

Step 2: Find the component of the gravitational force along the tunnel
Let \(x\) be the distance of the particle from the midpoint of the tunnel.
The distance from the center of the Earth to the particle is \(r=\sqrt{x^2+\left(\frac{R}{2}\right)^2}\).
The component of the gravitational force along the tunnel is:
\(F_x=F \cos \theta\)
where \(\cos \theta=\frac{x}{r}=\frac{x}{\sqrt{x^2+\left(\frac{R}{2}\right)^2}}\).
Substituting the values:
\(F_x=-\frac{G M m}{R^3} r \cdot \frac{x}{r}\)
\(F_x=-\frac{G M m}{R^3} x\)

Step 3: Determine the effective spring constant
Comparing \(F_x=-\frac{G M m}{R^3} x\) with \(F=-k x\), the effective spring constant is:
\(k=\frac{G M m}{R^3}\)

Step 4: Calculate the time period
The time period of the simple harmonic motion is:
\(T=2 \pi \sqrt{\frac{m}{k}}\)
\(T=2 \pi \sqrt{\frac{m}{\frac{G M m}{R^3}}}\)
\(T=2 \pi \sqrt{\frac{R^3}{G M}}\)
Since \(g=\frac{G M}{R^2}\), then \(G M=g R^2\).
Substituting \({G M}\) :
\(T=2 \pi \sqrt{\frac{R^3}{g R^2}}\)
\(T=2 \pi \sqrt{\frac{R}{g}}\)
The time period of the particle’s motion is \(2 \pi \sqrt{\frac{R}{g}}\).

Hint

(c) Two springs with spring constant \(K_1\) are connected in series.
For springs in series, the equivalent spring constant \(K_{e q}\) is given by \(\frac{1}{K_{e q}}=\frac{1}{K_1}+\frac{1}{K_2}\).
The time period of a spring-mass system is given by \(T=2 \pi \sqrt{\frac{m}{K}}\).

How to solve?
First find the equivalent spring constant for the series combination. Then, find the new time period and compare it with the original time period.

Step 1: Find the equivalent spring constant \(K_{\text {eq }}\)
For two springs in series:
\(\frac{1}{K_{e q}}=\frac{1}{K_1}+\frac{1}{K_1}\)
\(\frac{1}{K_{e q}}=\frac{2}{K_1}\)
\(K_{e q}=\frac{K_1}{2}\)
The new spring constant is \(\frac{1}{2}\) times the original.

Step 2: Find the new time period \(T^{\prime}\)
Original time period:
\(T=2 \pi \sqrt{\frac{m}{K_1}}\)
New time period:
\(\boldsymbol{T}^{\prime}=2 \pi \sqrt{\frac{m}{K_{e q}}}\)
\(\boldsymbol{T}^{\prime}=2 \pi \sqrt{\frac{m}{\frac{K_1}{2}}}\)
\(\boldsymbol{T}^{\prime}=2 \pi \sqrt{\frac{2 m}{K_1}}\)
\(\boldsymbol{T}^{\prime}=\sqrt{2} \cdot 2 \pi \sqrt{\frac{m}{K_1}}\)
\(T^{\prime}=\sqrt{2} T\)
The new time period is \(\sqrt{2}\) times the original.

The new spring constant is changed by a factor of \(\frac{1}{2}\) and the time period is changed by a factor of \(\sqrt{2}\).

Hint

(a) Time-displacement equation: \({Y}={A} \sin \left(\omega t+\phi_0\right)\)
At \(t=0\), displacement \(Y=\frac{A}{2}\)
The particle is moving along the negative -x-direction.

Velocity in SHM is given by the derivative of displacement with respect to time:
\(
v=\frac{d Y}{d t}=A \omega \cos \left(\omega t+\phi_0\right)
\)
The sign of velocity indicates the direction of motion.

How to solve?
Substitute the given values into the displacement and velocity equations and solve for \(\phi_0\).

Step 1: Find \(\phi_0\) using the displacement at \(t=0\)
Substitute \(t=0\) and \({Y}=\frac{{A}}{2}\) into the displacement equation:
\(\frac{A}{2}=A \sin \left(\omega \cdot 0+\phi_0\right)\)
\(\frac{A}{2}=A \sin \left(\phi_0\right)\)
\(\frac{1}{2}=\sin \left(\phi_0\right)\)

Possible solutions for \(\phi_0\) are \(\frac{\pi}{6}\) and \(\frac{5 \pi}{6}\).

Step 2: Determine the correct \(\phi_0\) using the direction of motion
Find the velocity equation:
\(v=\frac{d Y}{d t}=A \omega \cos \left(\omega t+\phi_0\right)\)
Substitute \({t}={0}\) into the velocity equation:
\(v=A \omega \cos \left(\phi_0\right)\)
Since the particle is moving in the negative x -direction, \(v<0\).
Check \(\phi_0=\frac{\pi}{6}\) :
\(v=A \omega \cos \left(\frac{\pi}{6}\right)=A \omega \frac{\sqrt{3}}{2}>0\) (incorrect)
Check \(\phi_0=\frac{5 \pi}{6}\) :
\(v=A \omega \cos \left(\frac{5 \pi}{6}\right)=A \omega\left(-\frac{\sqrt{3}}{2}\right)<0\) (correct)

The initial phase angle is \(\frac{5 \pi}{6}\).

Hint

(b) 

\(
\begin{aligned}
&\text { Due to parallel combination } K_{\text {eff }}=2 k+2 k=4 k\\
&\begin{aligned}
& \because T=2 \pi \sqrt{\frac{m}{k_{\text {eff }}}} \\
& =2 \pi \sqrt{\frac{m}{4 k}} \\
& T=\pi \sqrt{\frac{m}{k}}
\end{aligned}
\end{aligned}
\)

Hint

(a) The point ‘ \(A\) ‘ covers \(30^{\circ}\) in 0.1 sec .
Means, time period for covering area is:
For \(\pi / 6\), degree coverage of the area, we need time \(=0.1 \mathrm{sec}\)
For one degree time required is \(=0.1 /(\pi / 6)\)
For covering complete circle degrees are \(=2 \pi\)
For completing the circle time period required is \((T)=0.1 \times 6 \times 2 \pi / \pi\)
\(
{T}=1.2 \mathrm{sec} \dots(1)
\)
We know, that Angular frequency, \(\omega=2 \pi / T \dots(2)\)
Using equations (2) and (1) we get:
\(
\omega=2 \pi / 1.2
\)
We know restoring force \(F=m \omega^2 A\)
Then, Restoring force per unit mass \((F / m)=\omega^2 A\)
\(
\begin{aligned}
& \mathrm{F} / \mathrm{m}=(2 \pi / 1.2)^2 \times 0.36 \\
& \simeq 9.87 \mathbf{N}
\end{aligned}
\)

Hint

(b) What’s given in the problem
Length of the pendulum: \(l=2 \mathrm{~m}\)
Time period of the pendulum: \({T}=2 \mathrm{~s}\)

The formula for the time period of a simple pendulum is \(T=2 \pi \sqrt{\frac{l}{g}}\).

How to solve?
Solve the time period formula for g.

Step 1: Express \(g\) from the formula for the time period of a simple pendulum
Start with the formula:
\(T=2 \pi \sqrt{\frac{l}{g}}\)
Square both sides:
\(T^2=4 \pi^2 \frac{l}{g}\)
Solve for \(g\) :
\(g=\frac{4 \pi^2 l}{T^2}\)

Step 2: Substitute the given values into the equation
Substitute \(l=2 \mathrm{~m}\) and \(T=2 \mathrm{~s}\) :
\(g=\frac{4 \pi^2(2)}{(2)^2}\)

Simplify:
\(g=\frac{8 \pi^2}{4}\)
\(g=2 \pi^2\)

The acceleration due to gravity is \(2 \pi^2 \mathrm{~m} / \mathrm{s}^2\).

Hint

(c) No effect of inclination of plane on time period or frequency of oscillation.
Both springs are in parallel.
\(
\therefore \mathrm{k}_{e q}=k_1+k_2=k+k=2 k
\)
\(\Rightarrow\) Effective spring constant is 2 k
\(
\begin{aligned}
&T=2 \pi \sqrt{\frac{M}{k_{e q}}}\\
&\therefore T=2 \pi \sqrt{\frac{M}{2 k}}\left[\because k_{e q}=k+k\right]
\end{aligned}
\)
and frequency \((f)=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{2 k}{M}}\)

Hint

(d) As we know that the displacement of the simple harmonic motion we have;
\(x=A \sin \omega t \dots(1)\)
\(
\sin \omega t=\frac{x}{A} \dots(2)
\)
As we know that the velocity is equal to the rate of change of displacement per unit of time and it is written as;
\(
v=\frac{d x}{d t}
\)
Now, differentiate the equation (1) we have;
\(
\begin{aligned}
& {v}=\omega {A} \cos \omega {t} \dots(3) \\
& \cos \omega t=\frac{v}{\omega A} \dots(4)
\end{aligned}
\)
Now, squaring and adding equations (2) and (4) we have
\(
\frac{x^2}{A^2}+\frac{v^2}{\omega^2 A^2}=\sin ^2 \omega t+\cos ^2 \omega t=1
\)
Therefore,
\(
\frac{x^2}{A^2}+\frac{v^2}{\omega^2 A^2}=1 \dots(5)
\)
As we know that the equation for the ellipse is
\(
\frac{x^2}{A^2}+\frac{y^2}{B^2}=1 \dots(6)
\)
we can resemble the equation (5) and (6), therefore it is an ellipse.

Hint

(c) Given, Length of pendulum \(L=1.0 \mathrm{~m}\)
Error in length measurement \(\Delta L=1 \mathrm{~mm}=0.001 \mathrm{~m}\)
Time for one oscillation \(T=1.95 \mathrm{~s}\)
Error in time measurement \(\Delta T=0.01 \mathrm{~s}\)

The formula for the period of a simple pendulum is \(T=2 \pi \sqrt{\frac{L}{g}}\). The error propagation formula for \(g=\frac{4 \pi^2 L}{T^2}\) is \(\frac{\Delta g}{g}=\frac{\Delta L}{L}+2 \frac{\Delta T}{T}\). Percentage error is calculated as \(\frac{\Delta g}{g} \times 100 \%\).

How to solve
Calculate the percentage error in ‘g’ using the error propagation formula and the given values.

Step 1: Calculate the relative error in g
Use the formula for error propagation:
\(\frac{\Delta g}{g}=\frac{\Delta L}{L}+2 \frac{\Delta T}{T}\)
\(\frac{\Delta g}{g}=\frac{0.001}{1.0}+2 \frac{0.01}{1.95}\)
\(\frac{\Delta g}{g}=0.001+0.010256\)
\(\frac{\Delta g}{g}=0.011256\)

Step 2: Calculate the percentage error ing
Multiply the relative error by \(100 \%\) :
Percentage error \(=\frac{\Delta g}{g} \times 100 \%\)
Percentage error \(=0.011256 \times 100 \%\)
Percentage error \(=1.1256 \%\)

Step 3: Round the percentage error to two decimal places
Percentage error \(\approx 1.13 \%\)

Hint

(d) Conservation of Momentum:
When the mass \(m\) is added, the system’s momentum remains conserved because there are no external forces acting on the system.

Initial and Final Momentum:
Initial momentum (before adding \(m\) ) \(={MA} \omega\) where \(\omega\) is the angular frequency of the oscillation.
Final momentum (after adding \(m\) ) \(=(M+m) A^{\prime} \omega^{\prime}\) where \(A^{\prime}\) is the new amplitude and \(\omega^{\prime}\) is the new angular frequency.

Relating Angular Frequencies:
The angular frequency is related to the spring constant \(k\) and mass \(M\) by \(\omega=\sqrt{(k / M)}\) and \(\omega^{\prime}=\sqrt{(k /(M+m))}\).

Equating Momenta (Momentum of system remains conserved):
Setting initial momentum equal to final momentum: \(M A \sqrt{(k / M)}=(M+m) A^{\prime} \sqrt{(k /(M+m))}\).
Solving for New Amplitude (A’):
Simplifying the equation, we get \(A^{\prime}=A \sqrt{(M /(M+m))}\).

Hint

(d) The motion of the particle is described by the equation:
\(
y(t)=y_0 \sin ^2(\omega t)
\)
where \(y\) is the displacement from the lower end of the unstretched spring.
Using the trigonometric identity:
\(
\sin ^2(\theta)=\frac{1-\cos (2 \theta)}{2}
\)
we can rewrite the equation as:
\(
y(t)=y_0 \cdot \frac{1-\cos (2 \omega t)}{2}
\)
This simplifies to:
\(
y(t)=\frac{y_0}{2}-\frac{y_0}{2} \cos (2 \omega t)
\)
In the context of a mass-spring system, the angular frequency \(\omega\) is related to the spring constant \(k\) and the mass \(m\) by the formula:
\(
\omega=\sqrt{\frac{k}{m}}
\)
When the mass \(m\) is attached to the spring and released, it will stretch the spring to a new equilibrium position where the gravitational force equals the spring force:
\(
m g=k y
\)
where \(y\) is the displacement from the unstretched position.
At maximum amplitude, the total displacement from the equilibrium position can be expressed as:
\(
y_{\max }=y_0
\)
From the equilibrium condition, we can express the spring constant in terms of the maximum amplitude:
\(
k=\frac{m g}{y_0}
\)
\(
\begin{aligned}
&\text { Substituting } k \text { into the equation for } \omega \text { : }\\
&\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{m g}{m y_0}}=\sqrt{\frac{g}{y_0}}
\end{aligned}
\)
Thus, we arrive at the final expression for \(\omega\) :
\(
\omega=\sqrt{\frac{g}{2 y_0}}
\)

Hint

(d) Step 1: Understand the initial conditions
A block of mass \(m\) is attached to a massless spring and is performing oscillatory motion with an amplitude \(A\). At the mean position, the total energy of the system is purely kinetic.
Step 2: Write the expression for total energy
The total mechanical energy \({E}\) of the system when the block is at the mean position can be expressed as:
\(
E=\frac{1}{2} k A^2
\)
where \(k\) is the spring constant.
Step 3: Analyze the situation when half the mass is removed When half of the mass is removed while the block is passing through the mean position, the new mass becomes \(\frac{m}{2}\). Let the new velocity at the mean position be \(v^{\prime}\).
Step 4: Apply conservation of momentum
Using conservation of momentum at the moment the mass is halved:
\(
m v=\frac{m}{2} v^{\prime}
\)
From this, we can solve for \(v^{\prime}\) :
\(
v^{\prime}=2 v
\)
Step 5: Write the expression for kinetic energy after mass is halved
The kinetic energy at the mean position after the mass is halved is:
\(
\begin{gathered}
E^{\prime}=\frac{1}{2}\left(\frac{m}{2}\right)\left(v^{\prime}\right)^2=\frac{1}{2}\left(\frac{m}{2}\right)(2 v)^2 \\
=\frac{1}{2}\left(\frac{m}{2}\right)\left(4 v^2\right)=\frac{2 m v^2}{2}=m v^2
\end{gathered}
\)
Step 6: Relate the new kinetic energy to potential energy at the extreme position
At the extreme position, all the kinetic energy converts to potential energy:
\(
E^{\prime}=\frac{1}{2} k A^{\prime 2}
\)
Equating the two expressions for energy:
\(
m v^2=\frac{1}{2} k A^{\prime 2}
\)
Step 7: Relate the new kinetic energy to the initial total energy From the initial energy expression:
\(
\frac{1}{2} k A^2=\frac{1}{2} m v^2
\)
Thus, we can express \(m v^2\) in terms of \(A\) :
\(
m v^2=\frac{1}{2} k A^2
\)
Step 8: Substitute into the energy equation
Now substituting \(m v^2\) into the equation for \(E^{\prime}\) :
\(
\frac{1}{2} k A^2=\frac{1}{2} k A^{\prime 2}
\)
This simplifies to:
\(
A^{\prime 2}=2 A^2
\)
Taking the square root gives:
\(
A^{\prime}=\sqrt{2} A
\)
Step 9: Find the value of \(f\)
From the problem statement, we have \(A^{\prime}=f A\). Thus:
\(
f=\sqrt{2}
\)

Hint

(b)  (a) \({F}=ma\) and \({a}=-\omega^2 x\)
At \(\frac{3 T}{4}\) displacement zero \((x=0)\), so \({a}=0\)
\(
\therefore {F}=0
\)

(b) at \({t}={T}\), Particle is at extreme.
displacement \((x)=A \Rightarrow x\) maximum,
So acceleration is maximum.

(c) \({V}=\omega \sqrt{A^2-x^2}\)
at \({t}=\frac{T}{4}, {x}=0\)
\(
\therefore {V}={A} \omega={V}_{\max }
\)
(d)
\(
\begin{aligned}
&\mathrm{KE}=\mathrm{PE} \\
& \frac{1}{2} k\left(A^2-x^2\right)=\frac{1}{2} k x^2 \\
& \Rightarrow {~A}^2-{x}^2={x}^2 \\
& \Rightarrow {~A}^2=2 {x}^2 \\
& {~A}=\sqrt{2} {x} \\
& \Rightarrow {x}=\frac{A}{\sqrt{2}}={A} \cos \omega {t} \\
& \Rightarrow \cos \omega {t}=\frac{1}{\sqrt{2}} \\
& \Rightarrow \omega {t}=\frac{\pi}{4} \\
& \Rightarrow \frac{2 \pi}{T} \cdot t=\frac{\pi}{4} \\
& \Rightarrow {t}=\frac{T}{8}
\end{aligned}
\)

Hint

(b) Centripetal Force:
When the disc rotates, the mass experiences a centripetal force ( \(F=m \omega^2 r\) ) directed towards the center, where \(r\) is the radius of the circular path.

Spring Force:
This centripetal force is provided by the spring, which stretches or compresses to maintain the equilibrium.

Equilibrium Condition:
The spring force ( \({F}={kx}\), where \(x\) is the change in length) must equal the centripetal force ( \(m \omega^2 r\) ).

Derivation:
\(k x=m \omega^2 r\)
\(x=\left(m \omega^2 r\right) / k\)
Since the radius \(r\) is equal to the original length of the spring \(l\) plus the change in length \(x(r=l+x)\), we can substitute \(r\) :
\(x=\left(m \omega^2(l+x)\right) / k\)
\(k x=m \omega^2 l+m \omega^2 x\)
\(k x-m \omega^2 x=m \omega^2 l\)
\(x\left(k-m \omega^2\right)=m \omega^2 l\)
\(x=\left(m \omega^2 l\right) /\left(k-m \omega^2\right)\)
If we assume that the spring is not significantly stretched or compressed ( \(k \gg\) \(\left.m \omega^2\right)\), then \(x \approx\left(m \omega^2\right) / k\)
Therefore, the change in length \(x\) is approximately \(\left(\mathrm{m} \omega^2 l\right) / \mathrm{k}\).

Relative Change in Length:
The relative change in length is the change in length \((x)\) divided by the original length \(l\), which is \(\left({m} \omega^2 l\right) / \mathrm{k} / l={m} \omega^2 / \mathrm{k}\).

Hint

(a) 

\(
\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{l}{g}} \\
& \Rightarrow g=\frac{4 \pi^2 l}{T^2} \\
& \Rightarrow \frac{\Delta g}{g}=\frac{\Delta l}{l}+2 \frac{\Delta T}{T} \\
& =\frac{0.1}{25}+2 \frac{1}{50}=\frac{11}{250} \\
& \therefore \% \text { accuracy }=\frac{11}{250} \times 100 \%=4.40 \%
\end{aligned}
\)

Hint

(d) The equation of motion: \(y=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t) \mathrm{cm}\)
General form of SHM equation: \(y=A \sin (\omega t+\phi)\), where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant.
Trigonometric identity: \(a \sin x+b \cos x=R \sin (x+a)\), where \(R=\sqrt{a^2+b^2}\) and \(\alpha=\tan ^{-1}\left(\frac{b}{a}\right)\).
Relationship between angular frequency and time period: \(T=\frac{2 \pi}{\omega}\).

How to solve?
Rewrite the given equation in the standard form \(y=A \sin (\omega t+\phi)\).
Find the amplitude \(A\) and angular frequency \(\omega\).
Calculate the time period \(T\) using the formula \(T=\frac{2 \pi}{\omega}\).

Step 1: Rewrite the equation in the form \(R \sin (3 \pi t+\alpha)\)
Given equation: \(y=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)\)
Apply the trigonometric identity: \(a \sin x+b \cos x=R \sin (x+a)\)
Here, \(a=1, b=\sqrt{3}\), and \(x=3 \pi t\)
Calculate \(R: R=\sqrt{a^2+b^2}=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{1+3}=\sqrt{4}=2\)
Rewrite the equation: \(y=5 \cdot 2\left(\frac{1}{2} \sin 3 \pi t+\frac{\sqrt{3}}{2} \cos 3 \pi t\right)\)
\(
\begin{aligned}
& y=10\left(\sin 3 \pi t \cos \frac{\pi}{3}+\cos 3 \pi t \sin \frac{\pi}{3}\right) \\
& y=10 \sin \left(3 \pi t+\frac{\pi}{3}\right)
\end{aligned}
\)

Step 2: Identify the amplitude and angular frequency
Compare \(y=10 \sin \left(3 \pi t+\frac{\pi}{3}\right)\) with the standard form \(y=A \sin (\omega t+\phi)\)
Amplitude: \(A=10 \mathrm{~cm}\)
Angular frequency: \(\omega=3 \pi \mathrm{rad} / \mathrm{s}\)

Step 3: Calculate the time period
Use the formula: \(T=\frac{2 \pi}{\omega}\)
Substitute \(\omega=3 \pi\) : \(T=\frac{2 \pi}{3 \pi}=\frac{2}{3}\) s

The amplitude is 10 cm and the time period is \(\frac{2}{3} \mathrm{~s}\).

Hint

(b) When a system oscillates, the magnitude of restoring torque of the system is given by the formula:
\(
\tau=C \theta \dots(1)
\)

Where, \(\mathrm{C}=\) Constant that depends on system
The torque in terms of moment of inertia is given as:
\(
\tau=\mathrm{I} \alpha \dots(2)
\)
Where, \(\mathrm{I}\) = Moment of Inertia
\(\alpha=\) Angular acceleration
From equation (1) and equation (2),
\(
\begin{aligned}
& \Rightarrow \mathrm{I} \alpha=\mathrm{C} \theta \\
& \alpha=\frac{\mathrm{C}}{\mathrm{I}} \cdot \theta \dots(3)
\end{aligned}
\)
Now, the time period of oscillation of system is given by the formula:
\(
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{C}}}
\)
In given case, the magnitude of torque is given by the formula:
\(\tau=\) Force \(\times\) Perpendicular distance
\(
\begin{aligned}
&\tau=2 {kx} \times \frac{1}{2} \cos \theta\\
&\text { For small deflection, } \tau=\left(\frac{{kl}^2}{2}\right) \theta \dots(4)
\end{aligned}
\)
For small deflections, \(\sin \theta=\frac{x}{(1 / 2)} \approx \theta\)
\(
{x}=\frac{1}{2} \theta
\)
Also, \(\cos \theta=1\)
Comparing equation (4) and equation (1),
\(
\Rightarrow \mathrm{C}=\frac{{k}{l}^2}{2}
\)
Now, the equation (3) becomes,
\(
\begin{aligned}
& \Rightarrow \alpha=\frac{\left(k l^2 / 2\right)}{\left(\frac{1}{12} m l^2\right)} \theta \\
& \therefore \alpha=\frac{6 {k}}{{~m}} \theta
\end{aligned}
\)
Hence, time period of oscillation is:
\(
\begin{aligned}
&\begin{aligned}
& \Rightarrow T=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{C}}} \\
& \Rightarrow T=2 \pi \sqrt{\frac{\left(\frac{1}{12} m l^2\right)}{\left(\frac{\mathrm{kl}}{2}\right)}} \\
& \Rightarrow T=2 \pi \sqrt{\frac{m l^2}{12} \times \frac{2}{k l^2}} \\
& \therefore T=2 \pi \sqrt{\frac{\mathrm{~m}}{6 \mathrm{k}}}
\end{aligned}\\
&\text { Frequency of oscillation is given as: }\\
&\begin{aligned}
& \Rightarrow f=\frac{1}{T} \\
& \therefore f=\frac{1}{2 \pi} \sqrt{\frac{6 k}{m}}
\end{aligned}
\end{aligned}
\)

Hint

(d) Mass of planet: \(M_p=3 M_e\)
Diameter of planet: \({D}_{{p}}=3 {D}_{{e}}\), so radius of planet: \({R}_{{p}}=3 {R}_{{e}}\)
Period of pendulum on Earth: \({T}_e=2 \mathrm{~s}\)
Gravitational acceleration formula: \(\mathrm{g}=\frac{{G M}}{{R}^2}\)
Period of a simple pendulum: \(T=2 \pi \sqrt{\frac{L}{g}}\)
How to solve?
Find the ratio of gravitational acceleration on the planet to that on Earth, then use the relationship between the period and gravitational acceleration to find the period on the planet.

Step 1: Calculate the gravitational acceleration on Earth.
The gravitational acceleration on Earth is:
\(g_e=\frac{G M_e}{R_e^2}\)

Step 2: Calculate the gravitational acceleration on the planet.
The gravitational acceleration on the planet is:
\(g_p=\frac{G M_p}{R_p^2}\)
\(g_p=\frac{G\left(3 M_e\right)}{\left(3 R_e\right)^2}\)
\(g_p=\frac{3 G M_e}{9 R_e^2}\)
\(g_p=\frac{1}{3} \frac{G M_e}{R_e^2}\)
\(g_p=\frac{1}{3} g_e\)

Step 3: Find the ratio of the periods.
The period of a simple pendulum is given by:
\(T=2 \pi \sqrt{\frac{L}{g}}\)

Therefore, the ratio of the periods on the planet and Earth is:
\(\frac{T_p}{T_e}=\frac{2 \pi \sqrt{\frac{L}{g_p}}}{2 \pi \sqrt{\frac{L}{g_e}}}\)
\(\frac{T_p}{T_e}=\sqrt{\frac{g_e}{g_p}}\)
\(\frac{T_p}{T_e}=\sqrt{\frac{g_e}{\frac{1}{3} g_e}}\)
\(\frac{T_p}{T_e}=\sqrt{3}\)

Step 4: Calculate the period on the planet.
\(
\begin{aligned}
& T_p=T_e \sqrt{3} \\
& T_p=2 \sqrt{3} \mathrm{~S}
\end{aligned}
\)

Hint

(a)

\(
\text { Maximum kinetic energy }=\frac{1}{2} m \omega^2 A^2
\)
\(
\begin{aligned}
\omega & =\sqrt{\frac{g}{L}} \\
A & =L \theta
\end{aligned}
\)
\(
\begin{aligned}
& K E=\frac{1}{2} m \frac{g}{L} \times L^2 \theta^2, K E=\frac{1}{2} m g L \theta^2 \\
& K_1=\frac{1}{2} m g L \theta^2
\end{aligned}
\)
If Length Is Double
\(K_2=\frac{1}{2} m g(2 L)(\theta)^2\) [Here we are assuming angular amplitude is same]
\(
\begin{aligned}
& \frac{K_1}{K_2}=\frac{\frac{1}{2} m g l(\theta)^2}{\frac{1}{2} m g(2 L)(\theta)^2}=\frac{1}{2} \\
& K_2=2 K_1
\end{aligned}
\)

Hint

(b) Kinetic energy, \(\mathrm{k}=\frac{1}{2} m \omega^2 A^2 \cos ^2 \omega t\)
Potential energy, \(\mathrm{U}=\frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t\)
\(
\frac{k}{U}=\cot ^2 \omega t=\cot ^2 \frac{\pi}{90}(210)=\frac{1}{3}
\)

Hint

(d) The velocity \(({V})\) of a particle in SHM at a position \(x\) is given by:
\(
V=\omega \sqrt{A^2-x^2}
\)
\(
V=\omega \sqrt{(0.05)^2-(0.04)^2}=0.03 \omega
\)
The acceleration (a) of a particle in SHM at a position \(x\) is given by: \(a=-\omega^2 x\)
\(
a=-\omega^2(0.04)
\)
According to the problem, the magnitudes of velocity and acceleration are equal:
\(
0.03 \omega=0.04 \omega^2
\)
\(
\omega=\frac{0.03}{0.04}=\frac{3}{4} \mathrm{rad} / \mathrm{s}
\)
The time period \(({T})\) of SHM is given by:
\(
T=\frac{2 \pi}{\omega}
\)
Substituting the value of \(\omega\) : \(T=\frac{2 \pi}{\frac{3}{4}}=\frac{2 \pi \times 4}{3}=\frac{8 \pi}{3}\) seconds
The periodic time of the particle is: \(T=\frac{8 \pi}{3}\) seconds

Hint

(d)

\(
\begin{aligned}
&\text { Restoring force due to pressing the bottle with small amount } {x} \text {, }\\
&\begin{aligned}
& \mathrm{F}=-(\rho A x) g \\
& \Rightarrow \mathrm{ma}=-(\rho A x) g \\
& \Rightarrow \mathrm{a}=-\left(\frac{\rho A g}{m}\right) x \\
& \therefore \omega^2=\frac{\rho A g}{m}=\frac{\rho\left(\pi r^2\right) g}{m} \\
& \Rightarrow \omega=\sqrt{\frac{10^3 \times \pi \times\left(2.5 \times 10^{-2}\right)^2 \times 10}{310 \times 10^{-3}}}=7.90 \mathrm{rad} / \mathrm{s}
\end{aligned}
\end{aligned}
\)

Hint

(c) Mass of the rod: \({M}\)
Length of the rod: \(2 L\)
Distance of attached masses from the center: \(\frac{L}{2}\)
Reduction in oscillation frequency: 20%
Ipful information
Moment of inertia of a rod about its center: \(I_{\text {rod }}=\frac{1}{12} M L^2\)
Moment of inertia of a point mass: \(I_{\text {point }}=m r^2\)
Angular frequency of torsional oscillation: \(\omega=\sqrt{\frac{k}{I}}\), where \(k\) is the torsional constant and \(I\) is the moment of inertia.
How to solve?
Calculate the initial and final angular frequencies, set up an equation based on the given percentage reduction in frequency, and solve for the ratio \(\frac{m}{M}\).

Step 1: Calculate the initial moment of inertia \(I_1\) of the rod.
\(
\begin{aligned}
& I_1=\frac{1}{12} M(2 L)^2 \\
& I_1=\frac{1}{3} M L^2
\end{aligned}
\)
Step 2: Calculate the final moment of inertia \(I_2\) after adding the two masses.
\(
\begin{aligned}
& I_2=I_1+2 m\left(\frac{L}{2}\right)^2 \\
& I_2=\frac{1}{3} M L^2+\frac{1}{2} m L^2
\end{aligned}
\)
Step 3: Calculate the initial angular frequency \(\omega_1\).
\(
\begin{aligned}
& \omega_1=\sqrt{\frac{k}{I_1}} \\
& \omega_1=\sqrt{\frac{k}{\frac{1}{3} M L^2}} \\
& \omega_1=\sqrt{\frac{3 k}{M L^2}}
\end{aligned}
\)
Step 4: Calculate the final angular frequency \(\omega_2\).
\(
\begin{aligned}
& \omega_2=\sqrt{\frac{k}{I_2}} \\
& \omega_2=\sqrt{\frac{k}{\frac{1}{3} M L^2+\frac{1}{2} m L^2}}
\end{aligned}
\)
Step 5: Set up the equation based on the frequency reduction.
\(
\begin{aligned}
& \omega_2=\omega_1-0.2 \omega_1 \\
& \omega_2=0.8 \omega_1
\end{aligned}
\)
Step 6: Substitute the expressions for \(\omega_1\) and \(\omega_2\) into the equation.
\(
\sqrt{\frac{k}{\frac{1}{3} M L^2+\frac{1}{2} m L^2}}=0.8 \sqrt{\frac{3 k}{M L^2}}
\)
Step 7: Square both sides of the equation.
\(
\frac{k}{\frac{1}{3} M L^2+\frac{1}{2} m L^2}=0.64 \frac{3 k}{M L^2}
\)
Step 8: Simplify and solve for \(\frac{m}{M}\).
\(
\begin{aligned}
& \frac{1}{\frac{1}{3} M+\frac{1}{2} m}=\frac{1.92}{M} \\
& M=1.92\left(\frac{1}{3} M+\frac{1}{2} m\right) \\
& M=0.64 M+0.96 m \\
& 0.36 M=0.96 m \\
& \frac{m}{M}=\frac{0.36}{0.96} \\
& \frac{m}{M}=\frac{36}{96} \\
& \frac{m}{M}=\frac{3}{8} \\
& \frac{m}{M}=0.375
\end{aligned}
\)
The ratio of \(m\) to \(M\) is 0.375 .

Hint

(c) Total energy of particle \(=\frac{1}{2} k A^2\)
Potential energy \((U)=\frac{1}{2} k x^2\)
Kinetic energy \((K)=\frac{1}{2} k A^2-\frac{1}{2} k x^2\)
According to the question,
Potential energy = Kinetic energy
\(
\begin{aligned}
& \therefore \frac{1}{2} {kx}^2=\frac{1}{2} {kA}^2-\frac{1}{2} {kx}^2 \\
& \Rightarrow {kx}^2=\frac{1}{2} {kA}^2 \\
& \Rightarrow {x}= \pm \frac{A}{\sqrt{2}}
\end{aligned}
\)