Sample Exercises-II

Hint

It is given that:
Amplitude of simple harmonic motion, \(x=0.1 \mathrm{~m}\)
Time period of simple harmonic motion, \(T=0.314 \mathrm{~s}\)
Mass of the block, \(m=0.5 \mathrm{~kg}\)
Weight of the block, \(W=m g=0.5 \times 10=5 \mathrm{~kg} \quad\left(\because g=10 \mathrm{~ms}^{-2}\right)\)
Total force exerted on the block \(=\) Weight of the block + spring force

Periodic time of spring is given by,
\(
\begin{aligned}
& T=2 \pi \sqrt{\left(\frac{m}{k}\right)} \\
& \Rightarrow 0.314=2 \pi \sqrt{\left(\frac{0.5}{k}\right)} \\
& \Rightarrow k=200 N / m
\end{aligned}
\)
\(\therefore\) The force exerted by the spring on the block \((F)\) is, \(F=k x=200.0 \times 0.1=20 \mathrm{~N}\)
Maximum force \(=F+\) weight of the block
\(
=20+5=25 \mathrm{~N}
\)

Hint

It is given that:
Mass of the body, \(m=2 \mathrm{~kg}\)
Time period of the spring mass system, \(T=4 \mathrm{~s}\)
The time period for spring-mass system is given as,
\(
T=2 \pi \sqrt{\left(\frac{m}{k}\right)}
\)
where \(k\) is the spring constant.
On substituting the respective values, we get:
\(
\begin{aligned}
& \Rightarrow 4=2 \pi \sqrt{\frac{2}{k}} \\
& \Rightarrow 2=\pi \sqrt{\frac{2}{k}} \\
& \Rightarrow 4=\pi^2\left(\frac{2}{k}\right) \\
& \Rightarrow k=\frac{2 \pi^2}{4} \\
& =\frac{\pi^2}{2}=5 N / m
\end{aligned}
\)
As the restoring force is balanced by the weight, we can write:
\(
\begin{aligned}
F & =m g=k x \\
\Rightarrow x & =\frac{m g}{k}=\frac{2 \times 10}{5}=4
\end{aligned}
\)
\(\therefore\) Potential Energy \((U)\) of the spring is,
\(
\begin{aligned}
& U=\left(\frac{1}{2}\right) k x^2=\frac{1}{2} \times 5 \times 16 \\
& =5 \times 8=40 \mathrm{~J}
\end{aligned}
\)

Hint

It is given that:
Energy stored in the spring, \(E=5 \mathrm{~J}\)
Frequency of the mass-spring system, \(f=5\)
Extension in the length of the spring, \(x=25 \mathrm{~cm}=0.25 \mathrm{~m}\)
Time period,\(T=\frac{1}{5} s\)
Potential energy \((U)\) is given by,
\(
\begin{aligned}
& U=\frac{1}{2} k x^2 \\
& \Rightarrow \frac{1}{2} k x^2=5 \\
& \Rightarrow \frac{1}{2} k(0.25)^2=5 \\
& \Rightarrow k=160 N / m
\end{aligned}
\)
Time period of spring mass system is given by,
\(T=2 \pi \sqrt{\left(\frac{m}{k}\right)}\) where \(\mathrm{m}\) is the mass of the body hanged, and \(\mathrm{k}\) is the spring constant.
On substituting the respective values in the above expression, we get:
\(
\begin{aligned}
& \frac{1}{5}=2 \pi \sqrt{\left(\frac{m}{160}\right)} \\
& \Rightarrow m=0.16 \mathrm{~kg}
\end{aligned}
\)

Hint

(a) Consider the free body diagram.
Weight of the body, \(W=m g\)
Force, \(F=m a=m \omega^2 x\)
\(x\) is the small displacement of mass \(m\).
As normal reaction \(R\) is acting vertically in the upward direction, we can write:
\(
R+m \omega^2 x-m g=0 \dots(1)
\)
\(
\begin{aligned}
& \text { Resultant force }=m \omega^2 x=m g-R \\
& \Rightarrow m \omega^2 x=m\left(\frac{k}{M+m}\right) x \\
& =\frac{m k x}{M+m}
\end{aligned}
\)
Here,
\(
\omega=\sqrt{\left\{\frac{k}{M+m}\right\}}
\)
\(
\begin{aligned}
& \text { (b) } \mathrm{R}=m g-m \omega^2 x \\
& =m g-m \frac{k}{M+N} x \\
& =m g-\frac{m k x}{M+N}
\end{aligned}
\)
It can be seen from the above equations that, for \(R\) to be smallest, the value of \(m \omega^2 x\) should be maximum which is only possible when the particle is at the highest point.
(c) \(\mathrm{R}=m g-m \omega^2 x\)
As the two blocks oscillate together \(R\) becomes greater than zero.
When limiting condition follows,
i.e. \(R=0\)
\(
\begin{aligned}
& m g=m \omega^2 x \\
& x=\frac{m g}{m \omega^2}=\frac{m g \cdot(M+m)}{m k} \\
&
\end{aligned}
\)
Required maximum amplitude
\(
=\frac{g(M+m)}{k}
\)

Hint

(a) As it can be seen from the figure,
Restoring force \(=k x\)
Component of total weight of the two bodies acting vertically downwards \(=\left(m_1+m_2\right) g \sin \theta\)
At equilibrium,
\(
\begin{aligned}
& k x=\left(m_1+m_2\right) g \sin \theta \\
& \Rightarrow x=\frac{\left(m_1+m_2\right) g \sin \theta}{k}
\end{aligned}
\)
(b) It is given that:
Distance at which the spring is pushed,
\(
x_1=\frac{2}{k}\left(m_1+m_2\right) g \sin \theta
\)
As the system is released, it executes S.H.M.
where \(\omega=\sqrt{\frac{k}{m_1+m_2}}\)
When the blocks lose contact, \(P\) becomes zero. \(\quad\left(P\right.\) is the force exerted by mass \(m_1\) on mass \(\left.m_2\right)\)
\(
\therefore m_2 g \sin \theta=m_2 x_2 \omega^2=m_2 x_2 \times \frac{k}{m_1+m_2} \Rightarrow x_2=\frac{\left(m_1+m_2\right) g \sin \theta}{k}
\)
Therefore, the blocks lose contact with each other when the spring attains its natural length.
(c) Let \(v\) be the common speed attained by both the blocks.
Total compression \(=x_1+x_2\)
\(
\begin{aligned}
& \frac{1}{2}\left(m_1+m_2\right) v^2-0=\frac{1}{2} k\left(x_1+x_2\right)^2-\left(m_1+m_2\right) g \sin \theta\left(x+x_1\right) \\
& \Rightarrow \frac{1}{2}\left(m_1+m_2\right) v^2=\frac{1}{2} k\left(\frac{3}{k}\right)\left(m_1+m_2\right) g \sin \theta-\left(m_1+m_2\right) g \sin \theta\left(x_1+x_2\right) \\
& \Rightarrow \frac{1}{2}\left(m_1+m_2\right) v^2=\frac{1}{2}\left(m_1+m_2\right) g \sin \theta \times\left(\frac{3}{k}\right)\left(m_1+m_2\right) g \sin \theta \\
& \Rightarrow v=\sqrt{\left\{\frac{3}{k}\left(m_1+m_2\right)\right\}} g \sin \theta
\end{aligned}
\)

Hint

\(
\text { (a) Let us push the particle lightly against the spring C through displacement } x \text {. }
\)
As a result of this movement, the resultant force on the particle is \(k x\). The force on the particle due to springs \(\mathrm{A}\) and \(\mathrm{B}\) is \(\frac{k x}{\sqrt{2}}\).
Total Resultant force \(=k x+\sqrt{\left(\frac{k x}{\sqrt{2}}\right)^2+\left(\frac{k x}{\sqrt{2}}\right)^2}\) \(=k x+k x=2 k x\)
Acceleration is given by \(=\frac{2 k x}{m}\)
Time period \(=2 \pi \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}}\)
\(=2 \pi \sqrt{\frac{x}{2 k x / m}}\)
\(=2 \pi \sqrt{\frac{m}{2 k}}\)

Hint

\(
\text { As the particle is pushed against the spring } C \text { by the distance } x \text {, it experiences a force of magnitude } k x \text {. }
\)
If the angle between each pair of the springs is \(120^{\circ}\) then the net force applied by the springs \(A\) and \(B\) is given as,
\(
\sqrt{\left(\frac{k x}{2}\right)^2+\left(\frac{k x}{2}\right)^2+2\left(\frac{k x}{2}\right)\left(\frac{k x}{2}\right) \cos 120^{\circ}}=\frac{k x}{2}
\)
Total resultant force \((F)\) acting on mass \(m\) will be,
\(
\begin{aligned}
& F=k x+\frac{k x}{2}=\frac{3 k x}{2} \\
& \therefore a=\frac{F}{m}=\frac{3 k x}{2 m} \\
& \Rightarrow \frac{a}{x}=\frac{3 k}{2 m}=\omega^2 \\
& \Rightarrow \omega=\sqrt{\frac{3 k}{2 m}} \\
& \therefore \text { Time period }, T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{2 m}{3 k}}
\end{aligned}
\)

Hint

As the block of mass \(M\) is pulled, a net resultant force is exerted by the three springs opposing the motion of the block.
Now, springs \(k_2\) and \(k_3\) are in connected as a series combination.
Let \(k_4\) be the equivalent spring constant.
\(
\begin{aligned}
& \therefore \frac{1}{k_4}=\frac{1}{k_2}+\frac{1}{k_3}=\frac{k_2+k_3}{k_2 k_3} \\
& k_4=\frac{k_2 k_3}{k_2+k_3}
\end{aligned}
\)
\(k_4\) and \(k_1\) form a parallel combination of springs. Hence, equivalent spring constant \(k=k_1+k_4\).
\(
\begin{aligned}
& =\frac{k_2 k_3}{k_2+k_3}+k_1 \\
& =\frac{k_2 k_3+k_1 k_2+k_1 k_3}{k_2+k_3}
\end{aligned}
\)
\(\therefore\) Time peiod, \(T=2 \pi \sqrt{\frac{M}{k}}\)
\(
=2 \pi \sqrt{\frac{M\left(k_2+k_3\right)}{k_2 k_3+k_1 k_2+k_1 k_3}}
\)
(b) Frequency \((v)\) is given by,
\(
\begin{aligned}
& v=\frac{1}{T} \\
& =\frac{1}{2 \pi} \sqrt{\frac{k_2 k_3+k_1 k_2+k_1 k_3}{M\left(k_2+k_3\right)}}
\end{aligned}
\)
(c) Amplitude ( \(x\) ) is given by,
\(
x=\frac{F}{k}=\frac{F\left(k_2+k_3\right)}{k_1 k_2+k_2 k_3+k_1 k_3}
\)

Hint

All three spring attached to the mass \(M\) are in series.
\(\mathrm{k}_1, k_2, k_3\) are the spring constants.
Let \(k\) be the resultant spring constant.
\(
\begin{aligned}
& \frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3} \\
& \Rightarrow k=\frac{k_1 k_2 k_3}{k_1 k_2+k_2 k_3+k_3 k_1}
\end{aligned}
\)
Time period \((T)\) is given by,
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{M}{k}} \\
& =2 \sqrt{\frac{M\left(k_1 k_2+k_2 k_3+k_3 k_1\right)}{k_1 k_2 k_3}} \\
& =2 \sqrt{M\left(\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}\right)}
\end{aligned}
\)
As force is equal to the weight of the body,
\(
F=\text { weight }=M g
\)
Let \(x_1, x_2\), and \(x_3\) be the displacements of the springs having spring constants \(k_1, k_2\) and \(k_3\) respectively.
For spring \(k_1\)
\(
x_1=\frac{M g}{k_1}
\)
Similarly,\(x_2=\frac{M g}{k_2}\)
and \(x_3=\frac{M g}{k_3}\)
\(\therefore P E_1=\frac{1}{2} k_1 x_1^2\)
\(=\frac{1}{2} k_1\left(\frac{M g}{k_1}\right)^2\)
\(=\frac{1}{2} k_1 \frac{M^2 g^2}{k_1^2}\)
\(=\frac{1}{2} \frac{M^2 g^2}{k_1}=\frac{M^2 g^2}{2 k_1}\)
Similarly, \(P E_2=\frac{M^2 g^2}{2 k_2}\)
and \(P E_3=\frac{M^2 g^2}{2 k_3}\)

Hint

\(
\text { Let } l \text { be the extension in the spring when mass } m \text { is hung. }
\)
\(
T_1=k l=m g
\)
Let \(x\) be the extension in the string on applying a force \(F\). Then, the new value of tension \(T_2\) is given by,
\(
T_2=\mathrm{k}(\mathrm{x}+\mathrm{l})
\)
Driving force is the difference between tensions \(T_1\) and \(T_2\).
\(
\begin{aligned}
\therefore \text { Driving force } & =T_2-T_1=k(x+l)-k l \\
& =k x
\end{aligned}
\)
Acceleration, \(a=\frac{k x}{m}\)
Time period \((T)\) is given by,
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{\text { displacement }}{\text { Acceleration }}} \\
& =2 \pi \sqrt{\frac{x}{k x / m}}=2 \pi \sqrt{\frac{m}{k}}
\end{aligned}
\)

Hint

Let us try to solve the problem using energy method.
If \(\delta\) is the displacement from the mean position then, the initial extension of the spring from the mean position is given by, \(\delta=\frac{m g}{k}\)
Let \(x\) be any position below the equilibrium during oscillation.
Let \(v\) be the velocity of mass \(m\) and \(\omega\) be the angular velocity of the pulley.
If \(r\) is the radius of the pulley then
\(v=r \omega\)
As total energy remains constant for simple harmonic motion, we can write:
\(
\begin{aligned}
& \frac{1}{2} M v^2+\frac{1}{2} I \omega^2+\frac{1}{2} k\left[(x+\delta)^2-\delta^2\right]-M g x=\text { Constant } \\
& \Rightarrow \frac{1}{2} M v^2+\frac{1}{2} I \omega^2+\frac{1}{2} k x^2+k x d-M g x=\text { Constant } \\
& \Rightarrow \frac{1}{2} M v^2+\frac{1}{2} I\left(\frac{v^2}{r^2}\right)+\frac{1}{2} k x^2=\text { Constant }\left[\because \delta=\frac{M g}{k}\right]
\end{aligned}
\)
By taking derivatives with respect to \(t\), on both sides, we have:
\(
\begin{aligned}
& M v \cdot \frac{d v}{d t}+\frac{I}{r^2} v \cdot \frac{d v}{d t}+k x \frac{d x}{d t}=0 \\
& M v a+\frac{I}{r^2} v a+k x v=0\left(\because v=\frac{d x}{d t} \text { and } a=\frac{d v}{d t}\right) a\left(M+\frac{I}{r^2}\right)=-k x \Rightarrow \frac{a}{x}=\frac{k}{M+\frac{I}{r^2}}=\omega^2 T=\frac{2 \pi}{\omega} \\
& \Rightarrow T=2 \pi \sqrt{\frac{M+\frac{I}{r^2}}{k}}
\end{aligned}
\)

Hint

The centre of mass of the system should not change during simple harmonic motion.
Therefore, if the block \(m\) on the left hand side moves towards right by distance \(x\), the block on the right hand side should also move towards left by distance \(x\). The total compression of the spring is \(2 x\).
If \(v\) is the velocity of the block. Then
Using energy method, we can write:
\(
\begin{aligned}
& \frac{1}{2} k(2 x)^2+\frac{1}{2} m v^2+\frac{1}{2} m v^2=C \\
& \Rightarrow m v^2+2 k x^2=C
\end{aligned}
\)
By taking the derivative of both sides with respect to \(t\), we get:
\(
2 m v \frac{d v}{d t}+2 k \times 2 x \frac{d x}{d t}=0
\)
Putting \(v=\frac{d x}{d t} ;\) and \(a=\frac{d v}{d t}\) in above expression, we get
\(
\begin{aligned}
& m a+2 k x=0 \\
& \Rightarrow-\frac{a}{x}=\frac{2 k}{m}=\omega^2 \\
& \Rightarrow \omega=\sqrt{\frac{2 k}{m}} \\
& \Rightarrow \text { Time period }, T=2 \pi \sqrt{\left(\frac{m}{2 k}\right)}
\end{aligned}
\)

Hint

Let \(m\) is the mass of rectangular plate and \(\mathrm{x}\) is the displacement of the rectangular plate During the oscillation, the centre of mass does not change.
Driving force \((F)\) is given as,
\(F=m g \sin \theta\)
Comparing the above equation with \(\mathrm{F}=\mathrm{ma}\), we get:
\(
a=\frac{F}{m}=g \sin \theta
\)
For small values of \(\theta, \sin \theta\) can be taken as equal to \(\theta\).
Thus, the above equation reduces to:
\(a=g \theta=g\left(\frac{x}{L}\right)[\) Where \(\mathrm{g}\) and \(\mathrm{L}\) are constant.]
It can be seen from the above equation that, a \(a \propto x\).
Hence, the motion is simple harmonic.
Time period of simple harmonic motion \((T)\) is given by,
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{\text { displacement }}{\text { Acceleration }}} \\
& =2 \pi \sqrt{\frac{x}{g x / L}}=2 \pi \sqrt{\frac{L}{g}}
\end{aligned}
\)

Hint

It is given that: Amplitude of simple harmonic motion, \(x=0.1 \mathrm{~m}\)
Total mass of the system, \(M=3+1=4 \mathrm{~kg} \quad\) (when both the blocks move together)
Spring constant, \(k=100 \mathrm{~N} / \mathrm{m}\)
Time period of SHM \((T)\) is given by,
\(T=2 \pi \sqrt{\frac{M}{k}}\)
\(\text { On substituting the values of } M \text { and } k \text { in the above equation, we have: }\)
\(
T=2 \pi \sqrt{\frac{4}{100}}=\frac{2 \pi}{5} s
\)
Frequency of the motion is given by, \(\frac{1}{T}=\frac{5}{2 \pi} H z\)
Let \(v\) be the velocity of the \(1 \mathrm{~kg}\) block, at the mean position.
As kinetic energy is equal to the potential energy, we can write: \(\frac{1}{2} m v^2=\frac{1}{2} k x^2\)
where \(x=\) amplitude \(=0.1 \mathrm{~m}\)
Substituting the value of \(x\) in above equation and solving for \(v\), we get:
\(
\begin{aligned}
& \left(\frac{1}{2}\right) \times\left(1 \times v^2\right)=\left(\frac{1}{2}\right) \times 100(0.1)^2 \\
& v=1 m s^{-1} \ldots(1)
\end{aligned}
\)
When the \(3 \mathrm{~kg}\) block is gently placed on the \(1 \mathrm{~kg}\) block, the \(4 \mathrm{~kg}\) mass and the spring become one system. As a spring-mass system experiences an external force, momentum should be conserved.
Let \(V\) be the velocity of \(4 \mathrm{~kg}\) block.
Now, Initial momentum = Final momentum
\(\therefore 1 \times v=4 \times V\)
\(\Rightarrow V=\frac{1}{4} m / s\left[\right.\) As \(v=1 \mathrm{~ms}^{-1}\), from equation (1) \(]\)
Thus, at the mean position, two blocks have a velocity of \(\frac{1}{4} m s^{-1}\)
Mean value of kinetic energy is given as,
\(K E\) at mean position \(=\frac{1}{2} M V^2\)
\(
=\left(\frac{1}{2}\right) \times 4 \times\left(\frac{1}{4}\right)^2=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}
\)
At the extreme position, the spring-mass system has only potential energy.
\( E=\frac{1}{2} k \delta^2=\frac{1}{2} \times \frac{1}{4}\)
where \(\delta\) is the new amplitude.
\(
\begin{aligned}
& \therefore \frac{1}{4}=100 \delta^2 \\
& \delta=\sqrt{\left(\frac{1}{400}\right)} =0.05 \mathrm{~m}=5 \mathrm{~cm}
\end{aligned}
\)

Hint

According to the question, the collision is elastic and the surface is frictionless, therefore, when the left block A moves with speed \(v\) and collides with the right block \(B\), it transfers all the energy to the right block \(B\).
The left block A moves a distance \(x\) against the spring; the right block returns to the original position and completes half of the oscillation.
Therefore, the period of right block \(B\) will be, \(T=\frac{2 \pi \sqrt{\left(\frac{m}{k}\right)}}{2}=\pi \sqrt{\left(\frac{m}{k}\right)}\)
Right block B collides with left block A and comes to rest.
Let \(L\) be the distance moved by the block to return to its original position. The time taken is given by,
\(
\frac{L}{V}+\frac{L}{V}=2\left(\frac{L}{V}\right)
\)
Hence, time period of the periodic motion is,
\(
2 \frac{L}{V}+\pi \sqrt{\left(\frac{m}{k}\right)}
\)

Hint

Let \(t_1\) and \(t_2\) be the time taken by the particle to travel distances \(\mathrm{AB}\) and \(\mathrm{BC}\) respectively.
Acceleration for part \(\mathrm{AB}, a_1=g \sin 45^{\circ}\)
The distance travelled along \(A B\) is \(s_1\).
\(
\therefore s_1=\frac{0.1}{\sin 45^{\circ}}=2 m
\)
Let \(v\) be the velocity at point \(\mathrm{B}\), and \(u\) be the initial velocity.
Using the third equation of motion, we have:
\(
\begin{aligned}
& v^2-u^2=2 a_1 s_1 \\
& \Rightarrow v^2=2 \times g \sin 45^{\circ} \times \frac{0.1}{\sin 45^{\circ}}=2 \\
& \Rightarrow v=\sqrt{2} \mathrm{~m} / \mathrm{s} \\
& A s v=u+a_1 t_1 \\
& \therefore t_1=\frac{v-u}{a_1} \\
& =\frac{\sqrt{2}-0}{\frac{g}{\sqrt{2}}} \\
& =\frac{2}{g}=\frac{2}{10}=0.2 \sec \left(g=10 \mathrm{~ms}^{-2}\right)
\end{aligned}
\)
For the distance \(\mathrm{BC}\),
Acceleration, \(a_2=-g \sin 60^{\circ}\)
Initial velocity,\(u=\sqrt{2}\)
\(
\begin{aligned}
& v=0 \\
& \therefore \text { time period }, t_2=\frac{0-\sqrt{2}}{-\frac{g}{(3 \sqrt{2})}}=\frac{2 \sqrt{2}}{\sqrt{3} g} \\
& =\frac{2 \times(1.414)}{(1.732) \times 10}=0.163 \mathrm{~s}
\end{aligned}
\)
Thus, the total time period, \(t=2\left(t_1+t_2\right)=2(0.2+0.163)=0.73 \mathrm{~s}\)

Hint

Let \(x\) be the displacement of the uniform plate towards left.
Therefore, the centre of gravity will also be displaced through displacement \(x\).
At the displaced position,
\(
R_1+R_2=m g
\)
Taking moment about \(g\), we get:
\(
\begin{aligned}
& R_1\left(\frac{L}{2}-x\right)=R_2\left(\frac{L}{2}+x\right)=\left(M g-R_1\right)\left(\frac{L}{2}+x\right) \ldots (1)\\
& \therefore R_1\left(\frac{L}{2}-x\right)=\left(M g-R_1\right)\left(\frac{L}{2}+x\right) \\
& \Rightarrow R_1 \frac{L}{2}-R_1 x=M g \frac{L}{2}-R_1 x+M g x-R_1 \frac{L}{2} \\
& \Rightarrow R_1 \frac{L}{2}+R_1 \frac{L}{2}=M g\left(x+\frac{L}{2}\right) \\
& \Rightarrow R_1\left(\frac{L}{2}+\frac{L}{2}\right)=M g\left(\frac{2 x+L}{2}\right) \\
& \Rightarrow R_1 L=M g(2 x+L) \\
& \Rightarrow R_1=M g \frac{(L+2 x)}{2 L}
\end{aligned}
\)
Now,
\(
F_1=\mu R_1=\mu M g \frac{(L+2 x)}{2 L}
\)
Similarly ,
\(
F_2=\mu R_2=\mu M g \frac{(L-2 x)}{2 L}
\)
As \(F_1>F_2\), we can write:
\(
\begin{aligned}
& F_1-F_2=M a=\left(2 \mu \frac{M g}{L}\right) x \\
& \frac{a}{x}=2 \mu \frac{g}{L}=\omega^2 \\
& \Rightarrow \omega=\sqrt{\frac{2 \mu g}{L}}
\end{aligned}
\)
Time period \((T)\) is given by,
\(
T=2 \pi \sqrt{\frac{L}{2 \mu g}}
\)

Hint

It is given that:
Time period of the second pendulum, \(T=2 \mathrm{~s}\)
Acceleration due to gravity of a given place, \(g=\pi^2 \mathrm{~ms}^{-2}\)
The relation between time period and acceleration due to gravity is given by,
\(
T=2 \pi \sqrt{\left(\frac{l}{g}\right)}
\)
where \(l\) is the length of the second pendulum.
Substituting the values of \(T\) and \(g\), we get:
\(
\begin{aligned}
& \Rightarrow 2=2 \pi \sqrt{\left(\frac{l}{\pi^2}\right)} \\
& \Rightarrow \frac{1}{\pi}=\frac{\sqrt{l}}{\pi} \\
& \Rightarrow l=1 m
\end{aligned}
\)
Hence, the length of the pendulum is \(1 \mathrm{~m}\).

Hint

It is given that:
Angle made by the simple pendulum with the vertical, \(\theta=\left(\frac{\pi}{90}\right) \sin \left[\pi\left(s^{-1}\right) t\right]\) On comparing the above equation with the equation of \(\mathrm{S} . \mathrm{H} . \mathrm{M}\)., we get:
\(
\begin{aligned}
& \omega=\pi s^{-1} \\
& \Rightarrow \frac{2 \pi}{T}=\pi \\
& \therefore T=2 s
\end{aligned}
\)
Time period is given by the relation,
\(
\begin{aligned}
& T=2 \pi \sqrt{\left(\frac{l}{g}\right)} \\
& \Rightarrow 2=2 \pi \sqrt{\left(\frac{l}{\pi^2}\right)} \\
& \Rightarrow 1=\pi \frac{1}{\pi} \sqrt{(l)} \\
& \Rightarrow l=1 m
\end{aligned}
\)
Hence, the length of the pendulum is \(1 \mathrm{~m}\).

Hint

Given, Time period of the clock pendulum \(=2.04 \mathrm{~s}\)
The number of oscillations made by the pendulum in one day is calculated as \(\frac{\text { Number of seconds in one day }}{\text { time period of the pendulum in seconds }}=\frac{24 \times 3600}{2}=43200\)
In each oscillation, the clock gets slower by \((2.04-2.00)\) s, i.e., \(0.04 \mathrm{~s}\).
In one day, it is slowed by \(=43200 \times(0.04)\)
\(
=28.8 \mathrm{~min}
\)
Thus, the clock runs 28.8 minutes slow over 24 hours.

Hint

Let \(\mathrm{T}_1\) be the time period of the pendulum clock at a place where the acceleration due to gravity \(\left(g_1\right)\) is \(9.8 \mathrm{~ms}^{-2}\).
Let \(\mathrm{T}_1=2 \mathrm{~s}\)
\(
g_1=9.8 \mathrm{~ms}^{-2}
\)
Let \(T_2\) be the time period at the place where the pendulum clock loses 24 seconds during 24 hours.
Acceleration due to gravity at this place is \(\left(g_2\right)\)
\(
\begin{aligned}
& T_2=\frac{24 \times 3600}{\frac{(24 \times 3600-24)}{2}} \\
& =2 \times \frac{3600}{3599}
\end{aligned}
\)
As
\(
\begin{aligned}
& T \propto \frac{1}{\sqrt{g}} \\
& \therefore \frac{T_1}{T_2}=\sqrt{\left(\frac{g_2}{g_1}\right)} \\
& \Rightarrow \frac{g_2}{g_1}=\left(\frac{T_1}{T_2}\right)^2 \\
& \Rightarrow g_2=(9.8)\left(\frac{3599}{3600}\right)^2 \\
& =9.795 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)

Hint

It is given that:
Length of the pendulum, \(\mathrm{l}=5 \mathrm{~m}\)
Acceleration due to gravity, \(\mathrm{g}=9.8 \mathrm{~ms}^{-2}\)
Acceleration due to gravity at the moon, \(\mathrm{g}^{\prime}=1.67 \mathrm{~ms}^{-2}\)
(a) Time period \((T)\) is given by,
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{l}{g}} \\
& =2 \pi \sqrt{\frac{5}{9.8}} \\
& =2 \pi \sqrt{0.510}=2 \pi(0.71) \mathrm{s}
\end{aligned}
\)
i.e. the body will take \(2 \pi(0.7)\) seconds to complete an oscillation.
Now, frequency \((f)\) is given by,
\(
\begin{aligned}
& f=\frac{1}{T} \\
& \therefore f=\frac{1}{2 \pi(0.71)} \\
& =\frac{0.70}{\pi} H z
\end{aligned}
\)
(b) Let
\(g^{\prime}\) be the value of acceleration due to gravity at moon. Time period of simple pendulum at moon \(\left(T^{\prime}\right)\), is given as:
\(
T^{\prime}=2 \pi \sqrt{\left(\frac{l}{g^{\prime}}\right)}
\)
On substituting the respective values in the above formula, we get:
\(
T^{\prime}=2 \pi \sqrt{\frac{5}{1.67}}
\)
Therefore, frequency \(\left(f^{\prime}\right)\) will be,
\(
\begin{aligned}
& f^{\prime}=\frac{1}{T^{\prime}} \\
& =\frac{1}{2 \pi} \sqrt{\frac{1.67}{5}}=\frac{1}{2 \pi}(0.577) \\
& =\frac{1}{2 \pi \sqrt{3}} H z
\end{aligned}
\)

Hint

Let the speed of the bob of the pendulum at an angle \(\theta\) be \(v\).
Using the principle of conservation of energy between the mean and extreme positions, we get:
\(
\begin{aligned}
& \frac{1}{2} m v^2-0=m g l(1-\cos \theta) \\
& v^2=2 g l(1-\cos \theta) \dots(1)
\end{aligned}
\)
In a moving pendulum, the tension is maximum at the mean position, whereas it is minimum at the extreme position.
Maximum tension at the mean position is given by
\(
T_{\max }=m g+2 m g(1-\cos \theta)
\)
Minimum tension at the extreme position is given by
\(
T_{\min }=m g \cos \theta
\)
According to the question,
\(
T_{\max }=2 T_{\min }
\)
\(
\begin{aligned}
& \Rightarrow m g+2 m g-2 m g \cos \theta=2 m g \cos \theta \\
& \Rightarrow 3 m g=4 m g \cos \theta \\
& \Rightarrow \cos \theta=\frac{3}{4} \\
& \Rightarrow \theta=\cos ^{-1} \frac{3}{4}
\end{aligned}
\)

Hint

It is given that \(R\) is the radius of the concave surface.
Let \(N\) be the normal reaction force.
Driving force, \(F=m g \sin \theta\)
Comparing the expression for driving force with the expression, \(F=m a\), we get:
Acceleration, \(a=g \sin \theta\)
Since the value of \(\theta\) is very small,
\(\therefore \sin \theta \rightarrow \theta\)
\(\therefore\) Acceleration, \(a=g \theta\)
Let \(x\) be the displacement of the body from the mean position.
\(
\begin{aligned}
& \therefore \theta=\frac{x}{R} \\
& \Rightarrow a=g \theta=g\left(\frac{x}{R}\right) \\
& \Rightarrow\left(\frac{a}{x}\right)=\left(\frac{g}{R}\right) \\
& \Rightarrow a=x \frac{g}{R}
\end{aligned}
\)
As acceleration is directly proportional to the displacement. Hence, the body will execute S.H.M.
Time period \((T)\) is given by,
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{\text { displacement }}{\text { Acceleration }}} \\
& =2 \pi \sqrt{\frac{x}{g x / R}}=2 \pi \sqrt{\frac{R}{g}}
\end{aligned}
\)

Hint

Let \(\omega\) be the angular velocity of the system about the point of suspension at any time. The velocity of the ball rolling on a rough concave surface \(\left(v_C\right)\) is given by, \(v_C=(R-r) \omega\)
Also, \(v_{\mathrm{C}}=r \omega_1\)
where \(\omega_1\) is the rotational velocity of the sphere.
\(
\Rightarrow \omega_1=\frac{v_c}{r}=\left(\frac{R-r}{r}\right) \omega \cdots(1)
\)
As the total energy of a particle in S.H.M. remains constant,
\(
m g(R-r)(1-\cos \theta)+\frac{1}{2} m v_c^2+\frac{1}{2} I \omega_1^2=\text { constant }
\)
Substituting the values of \(v_{-} c\) and \(\omega_1\) in the above equation, we get:
\(
\begin{aligned}
& m g(R-r)(1-\cos \theta)+\frac{1}{2} m(R-r)^2 \omega^2+\frac{1}{2} m r^2\left(\frac{R-r}{r}\right) \omega^2=\text { constant }\left(\because I=m r^2\right) \\
& m g(R-r)(1-\cos \theta)+\frac{1}{2} m(R-r)^2 \omega^2+\frac{1}{5} m r^2\left(\frac{R-r}{r}\right) \omega^2=\text { constant } \\
& \Rightarrow g(R-r)(1-\cos \theta)+(R-r)^2 \omega^2\left[\frac{1}{2}+\frac{1}{5}\right]=\text { constant }
\end{aligned}
\)
Taking derivative on both sides, we get:
\(
g(R-r) \sin \theta \frac{d \theta}{d t}=\frac{7}{10}(R-r)^2 2 \omega \frac{d \omega}{d t}
\)
\(
\begin{aligned}
& \Rightarrow \mathrm{g} \sin \theta=2 \times\left(\frac{7}{10}\right)(\mathrm{R}-\mathrm{r}) \alpha\left(\because a=\frac{\mathrm{d} \omega}{\mathrm{dt}}\right) \\
& \Rightarrow \mathrm{g} \sin \theta=\left(\frac{7}{5}\right)(\mathrm{R}-\mathrm{r}) \alpha \\
& \Rightarrow \alpha=\frac{5 \mathrm{~g} \sin \theta}{7(\mathrm{R}-\mathrm{r})}=\frac{5 \mathrm{~g} \theta}{7(\mathrm{R}-\mathrm{r})}
\end{aligned}
\)
Therefore, the motion is S.H.M.
\(
\omega=\sqrt{\frac{5 g}{7(R-r)}}
\)
Time period is given by,
\(
\Rightarrow T=2 \pi \sqrt{\frac{7(R-r)}{5 g}}
\)

Hint

It is given that:
Length of the pendulum, \(l=40 \mathrm{~cm}=0.4 \mathrm{~m}\)
Radius of the earth, \(R=6400 \mathrm{~km}\)
Acceleration due to gravity on the earth’s surface, \(g=9.8 \mathrm{~ms}^{-2}\)
Let
\(g^{\prime}\) be the acceleration due to gravity at a depth of \(1600 \mathrm{~km}\) from the surface of the earth. Its value is given by,
\(g^{\prime}=g\left(1-\frac{d}{R}\right)\) where \(d\) is the depth from the earth surfce, \(\mathrm{R}\) is the radius of earth, and \(\mathrm{g}\) is acceleration due to gravity.
On substituting the respective values, we get: \(g^{\prime}=9.8\left(1-\frac{1600}{6400}\right)\)
\(
\begin{aligned}
& =9.8\left(1-\frac{1}{4}\right) \\
& =9.8 \times\left(\frac{3}{4}\right)=7.35 \mathrm{~ms}^{-2}
\end{aligned}
\)
Time period is given as,
\(
\begin{aligned}
& T=2 \pi \sqrt{\left(\frac{l}{g^{\prime}}\right)} \\
& \Rightarrow T=2 \pi \sqrt{\left(\frac{0.4}{7.35}\right)} \\
& \Rightarrow T=2 \times 3.14 \times 0.23 \\
& =1.465 \approx 1.47 \mathrm{~s}
\end{aligned}
\)

Hint

It is given that:
Length of the simple pendulum, \(l=1\) feet
Time period of simple pendulum, \(T=\frac{\pi}{3} s\) Acceleration due to gravity, \(\mathrm{g}=32 \mathrm{ft} / \mathrm{s}^2\)
Let \(a\) be the acceleration of the elevator while moving upwards.
Driving force \((f)\) is given by,
\(
f=m(g+a) \sin \theta
\)
Comparing the above equation with the expression, \(f=m a\), we get:
Acceleration, \(a=(g+a) \sin \theta=(g+a) \theta \quad\) (For small angle \(\theta, \sin \theta \rightarrow \theta\) )
\(=\frac{(g+a) x}{l}=\omega^2 x\) (From the diagram \(\left.\theta=\frac{x}{l}\right)\)
\(\Rightarrow \omega=\sqrt{\frac{(g+a)}{l}}\)
Time Period \((T)\) is given as,
\(
T=2 \pi \sqrt{\frac{l}{g+a}}
\)
On substituting the respective values in the above formula, we get:
\(
\begin{aligned}
& \frac{\pi}{3}=2 \pi \sqrt{\frac{1}{32+a}} \\
& \frac{1}{9}=4\left(\frac{1}{32+a}\right) \\
& \Rightarrow 32+a=36 \\
& \Rightarrow a=36-32=4 \mathrm{ft} / \mathrm{s}^2
\end{aligned}
\)

Hint

It is given that:
When the car is moving uniformly, time period of simple pendulum, \(T=4.0 \mathrm{~s}\)
As the accelerator is pressed, new time period of the pendulum, \(T^{\prime}=3.99 \mathrm{~s}\)
Time period of simple pendulum, when the car is moving uniformly on a horizontal road is given by,
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{l}{g}} \\
& \Rightarrow 4=2 \pi \sqrt{\frac{l}{g}}
\end{aligned}
\)
Let the acceleration of the car be \(a\).
The time period of pendulum, when the car is accelerated, is given by:
\(
\begin{aligned}
& T^{\prime}=2 \pi \sqrt{\frac{l}{\left(g^2+a^2\right)^{\frac{1}{2}}}} \\
& \Rightarrow 3.99=2 \pi \sqrt{\frac{l}{\left(g^2+a^2\right)^{\frac{1}{2}}}}
\end{aligned}
\)
Taking the ratio of \(T\) to \(T^{\prime}\), we get: \(\frac{T}{T^{\prime}}=\frac{4}{3.99}=\frac{\left(g^2+a^2\right)^{1 / 4}}{\sqrt{g}}\)
On solving the above equation for \(a\), we get:
\(
a=\frac{g}{10} m s^{-2}
\)

Hint

Given, Length of the long, light suspension wire, \(l=3 \mathrm{~cm}=0.03 \mathrm{~m}\)
Acceleration due to gravity, \(g=9.8 \mathrm{~ms}^{-2}\)
(a) Time Period \((T)\) is given by ,
\(
\begin{aligned}
& T=2 \pi \sqrt{\left(\frac{l}{g}\right)} \\
& =2 \pi \sqrt{\left(\frac{0.03}{9.8}\right)} \\
& =0.34 \text { second }
\end{aligned}
\)
(b) Velocity of merry-go-round, \(v=4 \mathrm{~ms}^{-1}\)
Radius of circle, \(r=2 m\)
As the lady sits on the merry-go-round, her earring experiences centripetal acceleration.
Centripetal acceleration (a) is given by,
\(
a=\frac{v^2}{r}=\frac{4^2}{2}=8 \mathrm{~m} / \mathrm{s}^2
\)
Resultant acceleration \((A)\) is given by ,
\(
\begin{aligned}
& A=\sqrt{\left(g^2+a^2\right)} \\
& =\sqrt{(96.04+64)} \\
& =12.65 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)
Time Period
\(
\begin{aligned}
& T=2 \pi \sqrt{\left(\frac{l}{A}\right)} \\
& =2 \pi \sqrt{\left(\frac{0.03}{12.65}\right)} \\
& =0.30 \text { second }
\end{aligned}
\)

Hint

It is given that the length of the rod is \(l\).
Let point \(A\) be the suspension point and point \(B\) be the centre of gravity.
Separation between the point of suspension and the centre of mass, \(l^{\prime}=\frac{l}{2}\)
\(\mathrm{AlsO}, h=\frac{l}{2}\)
Using the parallel axis theorem, the moment of inertia about \(A\) is given as,
\(
\begin{aligned}
& I=I_{C G}+m h^2 \\
& =\frac{m l^2}{12}+\frac{m l^2}{4}=\frac{m l^2}{3}
\end{aligned}
\)
the time period \((T)\) is given by,
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{I}{m g l^{\prime}}}=2 \pi \sqrt{\frac{I}{m g \frac{l}{2}}} \\
& =2 \pi \sqrt{\frac{2 m l^2}{3 m g l}}=2 \pi \sqrt{\frac{2 l}{3 g}}
\end{aligned}
\)
Let \(T^{\prime}\) be the time period of simple pendulum of length \(x\).
Time Period \(\left(T^{\prime}\right)\) is given by,\(T^{\prime}=2 \pi \sqrt{\left(\frac{x}{g}\right)}\)
As the time period of the simple pendulum is equal to the time period of the rod, \(T^{\prime}=T\)
\(
\begin{aligned}
& \Rightarrow \frac{2 l}{3 g}=\frac{x}{g} \\
& \Rightarrow x=\frac{2 l}{3}
\end{aligned}
\)

Hint

Suppose that the point is ‘ \(x\) ‘ distance from C.G.
Let \(m=\) mass of the disc. Radius \(=r\)
Here \(\ell=\mathrm{x}\)
M.I. about \(A=I_{\text {C.G. }}+m x^2=m r^2 / 2+m x^2=m\left(r^2 / 2+x^2\right)\)
\(
T=2 \pi \sqrt{\frac{1}{\mathrm{mg} \ell}}=2 \pi \sqrt{\frac{\mathrm{m}\left(\frac{r^2}{2}+x^2\right)}{\mathrm{mgx}}}=2 \pi \sqrt{\frac{\mathrm{m}\left(r^2+2 x^2\right)}{2 \mathrm{mgx}}}=2 \pi \sqrt{\frac{r^2+2 \mathrm{x}^2}{2 g x}} \dots(1)
\)
For \(\mathrm{T}\) is minimum \(\frac{\mathrm{dt}^2}{\mathrm{dx}}=0\)
\(
\therefore \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{T}^2=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{4 \pi^2 \mathrm{r}^2}{2 \mathrm{gx}}+\frac{4 \pi^2 2 \mathrm{x}^2}{2 \mathrm{gx}}\right)
\)
\(
\begin{aligned}
& \Rightarrow \frac{2 \pi^2 r^2}{g}\left(-\frac{1}{x^2}\right)+\frac{4 \pi^2}{g}=0 \\
& \Rightarrow-\frac{\pi^2 r^2}{g x^2}+\frac{2 \pi^2}{g}=0 \\
& \Rightarrow \frac{\pi^2 r^2}{g x^2}=\frac{2 \pi^2}{g} \Rightarrow 2 x^2=r^2 \Rightarrow x=\frac{r}{\sqrt{2}}
\end{aligned}
\)
So putting the value of equation (1)
\(
T=2 \pi \sqrt{\frac{r^2+2\left(\frac{r^2}{2}\right)}{2 g x}}=2 \pi \sqrt{\frac{2 r^2}{2 g x}}=2 \pi \sqrt{\frac{r^2}{g\left(\frac{r}{\sqrt{2}}\right)}}=2 \pi \sqrt{\frac{\sqrt{2} r^2}{g r}}=2 \pi \sqrt{\frac{\sqrt{2 r}}{g}}
\)

Hint

According to Energy equation, \(\mathrm{mg \ell}(1-\cos \theta)+(1 / 2) \mid \omega^2=\) const.
\(\operatorname{mg}(0.2)(1-\cos \theta)+(1 / 2) l \omega^2=C \dots(1)\).
Again, \(I=2 / 3 \mathrm{~m}(0.2)^2+\mathrm{m}(0.2)^2\)
\(
=m\left[\frac{0.008}{3}+0.04\right]
\)
\(=\mathrm{m}\left(\frac{0.1208}{3}\right) \mathrm{m}\). Where \(\mathrm{I} \rightarrow\) Moment of Inertia about the pt of suspension A
From equation
Differenting and putting the value of \(\mathrm{I}\) and 1 is
\(
\begin{aligned}
& \frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{mg}(0.2)(1-\cos \theta)+\frac{1}{2} \frac{0.1208}{3} \mathrm{~m} \omega^2\right]=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{C}) \\
& \Rightarrow \mathrm{mg}(0.2) \sin \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}+\frac{1}{2}\left(\frac{0.1208}{3}\right) \mathrm{m} 2 \omega \frac{\mathrm{d} \omega}{\mathrm{dt}}=0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow 2 \sin \theta=\frac{0.1208}{3} \alpha \text { [because, } g=10 \mathrm{~m} / \mathrm{s}^2 \text { ] } \\
& \Rightarrow \frac{\alpha}{\theta}=\frac{6}{0.1208}=\omega^2=58.36 \\
& \Rightarrow \omega=7.3 . \text { So } T=\frac{2 \pi}{\omega}=0.89 \mathrm{sec} .
\end{aligned}
\)
For simple pendulum \(\mathrm{T}=2 \pi \sqrt{\frac{0.19}{10}}=0.86 \mathrm{sec}\).
\(
\% \text { more }=\frac{0.89-0.86}{0.89}=0.3 .
\)
\(\therefore\) It is about \(0.3 \%\) larger than the calculated value.

Hint

\(
\text { M.I. of the centre of the disc. }=\mathrm{mr}^2 / 2
\)
\(
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{k}}}=2 \pi \sqrt{\frac{\mathrm{mr}^2}{2 \mathrm{~K}}} \text { [where } \mathrm{K}=\text { Torsional constant] }
\)
\(
\mathrm{T}^2=4 \pi^2 \frac{\mathrm{mr}^2}{2 \mathrm{~K}}=2 \pi^2 \frac{\mathrm{mr}^2}{\mathrm{~K}}
\)
\(
\begin{aligned}
& \Rightarrow 2 \pi^2 \mathrm{mr}^2=\mathrm{KT}^2 \Rightarrow \mathrm{K}=\frac{2 \mathrm{mr}^2 \pi^2}{\mathrm{~T}^2} \\
& \therefore \text { Torsional constant } \mathrm{K}=\frac{2 \mathrm{mr}^2 \pi^2}{\mathrm{~T}^2}
\end{aligned}
\)

Hint

It is given that the mass of both the balls is \(m\) and they are connected to each other with the help of a light rod of length \(L\).
Moment of inertia of the two-ball system \((I)\) is given by,
\(
I=2 m\left(\frac{L}{2}\right)^2=\frac{m L^2}{2}
\)

Torque \((\tau)\), produced at any given position \(\theta\) is given as: \(\tau=k \theta\)
\(\Rightarrow\) Work done during the displacement of the system from 0 to \(\theta_0\) will be,
\(
W=\int_0^{\theta_0} \mathrm{k} \theta d \theta=\frac{k \theta_0^2}{2}
\)
On applying the work-energy theorem, we get:
\(
\begin{aligned}
& \frac{1}{2} I \omega^2-0=\text { work done }=\frac{k \theta_0^2}{2} \\
& \therefore \omega^2=\frac{k \theta_0^2}{I}=\frac{k \theta_0^2}{m L^2}
\end{aligned}
\)

From the free-body diagram of the rod, we can write:
\(
\begin{aligned}
& \text { Force, } T_2=\sqrt{\left(m \omega^2 L\right)^2+(m g)^2} \\
& =\sqrt{\left(m \frac{k \theta_0^2}{m L^2} \times L\right)^2+m^2 g^2} \\
& =\sqrt{\frac{k^2 \theta_0^4}{L^2}+m^2 g^2}
\end{aligned}
\)

Hint

It is given that a particle is subjected to two S.H.M.s of same time period in the same direction.
Amplitude of first motion, \(A_1=3 \mathrm{~cm}\)
Amplitude of second motion, \(A_2=4 \mathrm{~cm}\)
Let \(\phi\) be the phase difference.
The resultant amplitude \((R)\) is given by,
\(
R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}
\)
(a) When \(\phi=0^{\circ}\)
\(
\begin{aligned}
& R=\sqrt{\left(3^2+4^2+(2)(3)(4) \cos 0^{\circ}\right)} \\
& =7 \mathrm{~cm}
\end{aligned}
\)
(b) When \(\phi=60^{\circ}\)
\(
\begin{aligned}
& R=\sqrt{3^2+4^2+(2)(3)(4) \cos 60^{\circ}} \\
& =\sqrt{37}=6.1 \mathrm{~cm}
\end{aligned}
\)
(c) When \(\phi=90^{\circ}\)
\(
\begin{aligned}
& R=\sqrt{\left(3^2+4^2+(2)(3)(4) \cos 90^{\circ}\right)} \\
& =\sqrt{25}=5 \mathrm{~cm}
\end{aligned}
\)

Hint

\(
\text { Three SHMs of equal amplitudes ‘A’ and equal time periods in the same direction combine. }
\)
The vectors representing the three SHMs are shown it the figure.
Using vector method,
Resultant amplitude \(=\) Vector sum of the three vectors
\(
=\mathrm{A}+\mathrm{A} \cos 60^{\circ}+\mathrm{A} \operatorname{cso} 60^{\circ}=\mathrm{A}+\mathrm{A} / 2+\mathrm{A} / 2=2 \mathrm{~A}
\)
So the amplitude of the resultant motion is \(2 \mathrm{~A}\).

Hint

Given are the equations of motion of a particle:
\(
\begin{aligned}
& x_1=2.0 \sin 100 \pi t \\
& x_2=2.0 \sin \left(120 \pi t+\frac{\pi}{3}\right)
\end{aligned}
\)
The Resultant displacement \((x)\) will be
\(
\begin{aligned}
& x=x_1+x_2 \\
& =2\left[\sin (100 \pi t)+\sin \left(120 \pi t+\frac{\pi}{3}\right)\right] \\
& \text { (a) At } t=0.0125 \mathrm{~s} \\
& x=2\left[\sin (100 \pi \times 0.0125)+\sin \left(120 \pi \times 0.0125+\frac{\pi}{3}\right)\right] \\
& =2\left[\sin \left(\frac{5 \pi}{4}\right)+\sin \left(\frac{3 \pi}{2}+\frac{\pi}{3}\right)\right] \\
& =2[(-0.707)+(-0.5)] \\
& =2 \times(-1.207)=-2.41 \mathrm{~cm}
\end{aligned}
\)
(b) At \(t=0.025 \mathrm{~s}\)
\(
\begin{aligned}
& x=2\left[\sin (100 \pi \times 0.025)+\sin \left(120 \pi \times 0.025+\frac{\pi}{3}\right)\right]=2\left[\sin \left(\frac{10 \pi}{4}\right)+\sin \left(3 \pi+\frac{\pi}{3}\right)\right] \\
& =2[1+(-0.866)] \\
& =2 \times(0.134)=0.27 \mathrm{~cm}
\end{aligned}
\)

Hint

Given:
Equation of motion along \(\mathrm{X}\) axis, \(x=x_0 \sin \omega t\)
Equation of motion along \(Y\) axis, \(s=s_0 \sin \omega t\)
Angle between the two motions, \(\theta 45^{\circ}\)
Resultant motion \((R)\) will be,
\(
\begin{aligned}
& R=\sqrt{\left(x^2+s^2+2(x)(s) \cos 45^{\circ}\right)} \\
& =\sqrt{\left\{x_0^2 \sin \omega t+s_0^2 \sin \omega t+2 x_0 s_0 \sin ^2 \omega t\left(\frac{1}{\sqrt{2}}\right)\right\}}=\left[x_0^2+s_0^2+\sqrt{2 x_0 s_0}\right]^{1 / 2} \sin \omega t
\end{aligned}
\)
Hence, the resultant amplitude is
\(
\left[x_0^2+s_0^2+\sqrt{2 x_0 s_0}\right]^{1 / 2}
\)