Past NEET Papers

Hint

Displacement equation of SHM of frequency ‘ \(n\) ‘ \(x=A \sin (\omega t)=A \sin (2 \pi n t)\)
Now,
Potential energy \(U=\frac{1}{2} \mathrm{kx}^2=\frac{1}{2} \mathrm{KA}^2 \sin ^2(2 \pi \mathrm{nt})\)
\(
=\frac{1}{2} \mathrm{kA}^2\left[\frac{1-\cos (2 \pi(2 \mathrm{n}) \mathrm{t})}{2}\right]
\)
So frequency of potential energy \(=2 n\)

Hint

Given, the mass of suspended, \(m=2 \mathrm{~kg}\) The spring is stretched, \(x=5 \mathrm{~cm}=0.05\) \(\mathrm{m}\)
The constant force applied on the spring, \(F=10 \mathrm{~N}\)
As we know that, spring force,
\(
\begin{aligned}
F & =k x \Rightarrow 10 \mathrm{~N}=k(0.05 \mathrm{~m}) \\
\Rightarrow \quad k & =200 \mathrm{~N} / \mathrm{m}
\end{aligned}
\)
Now, time period of the oscillation,
\(
T=2 \pi \sqrt{\frac{m}{k}} \Rightarrow T=2 \pi \sqrt{\frac{2}{200}}
\)
Time period, \(T=0.628 \mathrm{~s}\)

Hint

(c)
\(\sin \omega t\) and \(\cos \omega t\), both are periodic
function of period \(\frac{2 \pi}{\omega}\).
We know that, the sum of two periodic
functions is also a periodic function,
hence, \(\sin \omega t+\cos \omega t\) represents
periodic motion.

Hint

(c) Displacement equation of a SHM
\(
y=A \sin (\omega t+\phi)
\)
\(\therefore\) Velocity, \(v=\frac{d y}{d t}=A \omega \cos (\omega t+\phi)\)
Acceleration, \(a=\frac{d v}{d t}\)
or, \(a=-A \omega^2 \sin (\omega t+\phi)\)
\(
\therefore a=A \omega^2 \sin (\omega t+\phi+\pi)
\)
Hence, the phase difference between displacement and acceleration is \(\pi\).

Hint

In a simple harmonic motion (SHM) the particle oscillates about its mean position on a straight line.
The particle moves from its mean position (0) to an extreme position (P) and then return to its mean position covering same distance of \(A\).
Then by the conservative force, it is moved in the opposite direction to a point \(Q\) by distance \(A\) and then back to the mean position covering a distance of \(A\). This comprises of one time period as shown below

Hence, in one time period it covers a distance of
\(\)
\begin{aligned}
x & =O P+P O+O Q+Q O \\
& =A+A+A+A=4 A
\end{aligned}
\(\)

Hint

(c) Displacement of the particle in one complete vibration is zero, so, the average velocity in one complete vibration will be
\(
=\frac{\text { Displacement }}{\text { Time interval }}=\frac{y_f-y_i}{T}=0
\)

Hint

The displacement of given particle is
\(
y=A_0+A \sin \omega t+B \cos \omega t \dots(i)
\)
The general equation of \(\mathrm{SHM}\) can be given as
\(
x=a \sin \omega t+b \cos \omega t \dots(ii)
\)
So, from Eqs. (i)and (ii), we can say that \(A_0\) be the value of mean position, at which \(y=0\).
\(\therefore\) Amplitude, \(R=\sqrt{A^2+B^2+2 A B \cos \theta}\)
As two functions sine and cosine have phase shift to \(90^{\circ}\).
\(
\therefore R=\sqrt{A^2+B^2} \quad\left[\because \cos 90^{\circ}=0\right]
\)

Hint

Let \(O\) be the centre of circle, then at \(t=0\), the displacement \(y\) is maximum and have value \(3 \mathrm{~m}\).

As the general equation of displacement of a particle will be in the form
\(
y=A \cos \omega t
\)
Here, \(A=3 \mathrm{~m}\)
Then, \(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{4} \quad\) [given, \(T=4 \mathrm{~s}\) ]
\(
\begin{aligned}
& =\frac{\pi}{2} \\
\therefore \quad y & =3 \cos \left(\frac{\pi}{2} t\right) \text { (in metre) }
\end{aligned}
\)

Hint

(d) At surface of earth time taken in falling \(h\) distance.
\(
t=\sqrt{\frac{2 h}{g}}
\)
and \(T=2 \pi \sqrt{\frac{l}{g}}\)
Given \(t=2 \mathrm{~T}\)
\(
\frac{t}{T}=2
\)
For surface of other planet
\(
g^{\prime}=\frac{g}{2}
\)
Time taken in falling \(\mathrm{h}\) distance
\(
t^{\prime}=\sqrt{\frac{2 h}{g^{\prime}}}=\sqrt{2} t
\)
and \(T^{\prime}=2 \pi \sqrt{\frac{l}{g^{\prime}}}=\sqrt{2} T\)
Here \(\frac{t^{\prime}}{T^{\prime}}=\frac{\sqrt{2} t}{\sqrt{2} T}=2\)
\(
\therefore \quad t^{\prime}=2 T^{\prime}
\)

Hint

(b) From the question, acceleration, \(a=20 \mathrm{~m} / \mathrm{s}^2\), and displacement, \(\mathrm{y}=5 \mathrm{~m}\)
\(
\begin{aligned}
& |\mathrm{a}|=\omega^2 \mathrm{y} \\
\Rightarrow & 20=\omega^2(5) \\
\Rightarrow & \omega=2 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
The time period of the pendulum,
\(
\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi \mathrm{~s}
\)

Hint

Thinking Process Magnitude of velocity of particle when it is at displacement \(x\) from mean position
\(
=\omega \sqrt{A^2-x^2}
\)
Also, magnitude of acceleration of particle in SHM
\(
=\omega^2 x
\)
Given, when \(x=2 \mathrm{~cm}\)
According to question, magnitude of velocity \(=\) acceleration
\(
\begin{aligned}
& \quad|\mathbf{v}|=|\mathbf{a}| \Rightarrow \omega \sqrt{A^2-x^2}=\omega^2 x \\
& \Rightarrow \quad \omega=\frac{\sqrt{A^2-x^2}}{x}=\frac{\sqrt{9-4}}{2} \\
& \Rightarrow \text { Angular velocity } \omega=\frac{\sqrt{5}}{2} \\
& \therefore \text { Time period of motion } \\
& \quad T=\frac{2 \pi}{\omega}=\frac{4 \pi}{\sqrt{5}} \mathrm{~s}
\end{aligned}
\)

Hint

(c) Let \(\ell\) be the complete length of the spring. Length when cut in ratio, \(1: 2: 3\) are \(\frac{\ell}{6}, \frac{\ell}{3}\) and \(\frac{\ell}{2}\)
Spring constant \((\mathrm{k}) \propto \frac{1}{\text { length }(\ell)}\)
Spring constant for given segments \(\mathrm{k}_1=6 \mathrm{k}, \mathrm{k}_2=3 \mathrm{k}\) and \(\mathrm{k}_3=2 \mathrm{k}\)
When they are connected in series
\(
\begin{aligned}
& \frac{1}{\mathrm{k}^{\prime}}=\frac{1}{6 \mathrm{k}}+\frac{1}{3 \mathrm{k}}+\frac{1}{2 \mathrm{k}} \\
& \Rightarrow \frac{1}{\mathrm{k}^{\prime}}-\frac{6}{6 \mathrm{k}}
\end{aligned}
\)
\(\therefore\) Force constant \(\mathrm{k}^{\prime}=\mathrm{k}\)
And when they are connected in parallel \(\mathrm{k}^{\prime \prime}=6 \mathrm{k}+3 \mathrm{k}+2 \mathrm{k}\)
\(
\Rightarrow \mathrm{k}^{\prime \prime}=11 \mathrm{k}
\)
Then the ratios
\(
\frac{\mathrm{k}^{\prime}}{\mathrm{k}^{\prime \prime}}=\frac{1}{11} \text { i.e., } \mathrm{k}^{\prime}: \mathrm{k}^{\prime \prime}=1: 11
\)

Note: If a spring of force constant \(K\) is divided into \(n\) equal parts then the spring constant of each part will become \(n k\) and if these \(n\) parts connected in parallel then \(k_{\mathrm{eff}}=n^2 k\).

Hint

As we know that
Time period, \(T=2 \pi \sqrt{\frac{m}{k}}\)

Case I:
\(
T_1=2 \pi \sqrt{\frac{m}{k}} \dots(i)
\)

Case II: When the mass mis increased by
\(1 \mathrm{~kg}\), then \(=m+1\)
\(T_2=2 \pi \sqrt{\frac{m+1}{k}} \dots(ii)\)
From Eqs. (ii) and (i), we get
\(
\begin{aligned}
\frac{T_2}{T_1} & =\sqrt{\frac{m+1}{m}} \\
\Rightarrow \frac{5}{3} & =\sqrt{\frac{m+1}{m}} \Rightarrow \frac{25}{9}=\frac{m+1}{m} \\
\Rightarrow \frac{25}{9} & =1+\frac{1}{m} \Rightarrow \frac{1}{m}=\frac{16}{9} \\
\therefore & m=\frac{9}{16} \mathrm{~kg}
\end{aligned}
\)

Hint

The two displacements equations are \(\mathrm{y}_1=\mathrm{a} \sin (\omega \mathrm{t})\)
and \(\mathrm{y}_2=\mathrm{b} \cos (\omega \mathrm{t})=\mathrm{b} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}\right)\)
\(
\begin{aligned}
\mathrm{y}_{\mathrm{eq}} & =\mathrm{y}_1+\mathrm{y}_2 \\
& =\mathrm{a} \sin \omega \mathrm{t}+\mathrm{b} \cos \omega \mathrm{t}
\end{aligned}
\)
\(
=\mathrm{a} \sin \omega \mathrm{t}+\mathrm{b} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}\right)
\)
Since the frequencies for both SHMs are same, resultant motion will be SHM.

Now amplitude, \(A_{e q}=\sqrt{a^2+b^2+2 a b \cos \frac{\pi}{2}}\)
\(
\Rightarrow A_{e q}=\sqrt{a^2+b^2}
\)

Hint

(b) As we know, for particles undergoing SHM,
\(
\begin{aligned}
& V=\omega \sqrt{A^2-X^2} \\
& V_1^2=\omega^2\left(A^2-x_1^2\right) \\
& V_2^2=\omega^2\left(A^2-x_2^2\right)
\end{aligned}
\)
Substracting we get,
\(
\begin{aligned}
& \frac{V_1^2}{\omega^2}+x_1^2=\frac{V_2^2}{\omega^2}+x_2^2 \\
& \Rightarrow \quad \frac{V_1^2-V_2^2}{\omega^2}=x_2^2-x_1^2 \\
& \Rightarrow \quad \omega=\sqrt{\frac{V_1^2-V_2^2}{x_2^2-x_1^2}} \\
& \Rightarrow \quad T=2 \pi \sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}}
\end{aligned}
\)

Hint

For a particle executing SHM, we have maximum acceleration,
\(
\alpha=A \omega^2 \dots(i)
\)
where, \(A\) is maximum amplitude and \(\omega\) is angular velocity of a particle.
Maximum velocity, \(\beta=A \omega \dots(ii)\)
Dividing Eq. (i) by Eq. (ii), we get
\(
\frac{\alpha}{\beta}=\frac{A \omega^2}{A \omega} \Rightarrow \frac{\alpha}{\beta}=\omega=\frac{2 \pi}{T}
\)
i.e. \(T=\frac{2 \pi \beta}{\alpha}\)
Thus, its time period of vibration,
\(
T=\frac{2 \pi \beta}{\alpha}
\)

Hint

\(
\begin{aligned}
& A s, x=A \cos \omega t \\
& \therefore v=\frac{d x}{d t}=-A \omega \sin \omega t \dots(i) \\
& \text { and } a=\frac{d^2 x}{d t^2}=-A \omega^2 \cos \omega t \dots(ii)
\end{aligned}
\)
We can find the correct graph by putting different values of \(t\) in Eq. (ii).
At \(t=0, \quad a=-A \omega^2\)
At \(t=\frac{T}{4}, a=-A \omega^2 \cos \left(\frac{2 \pi}{T} \times \frac{T}{4}\right)=0\)
At \(t=\frac{T}{2}, a=-A \omega^2 \cos \left(\frac{2 \pi}{T} \times \frac{T}{2}\right)\)
\(=-A \omega^2 \cos \pi=+A \omega^2\)
At \(t=\frac{3 T}{4}, a=-A \omega^2 \cos \left(\frac{2 \pi}{T} \times \frac{3 T}{4}\right)=0\)
At \(t=T, a=-A \omega^2 \cos \left(\frac{2 \pi}{T} \times T\right)=-A \omega^2\)
This condition is represented by graph in option (c).

Hint

(d) Explanation:
\(
x=a \sin \omega t \text { or } \frac{x}{a}=\sin \omega t \dots(i)
[latex]
Velocity, [latex]v=\frac{d x}{d t}=a \omega \cos \omega t[latex]
[latex]
\frac{v}{a \omega}=\cos \omega t \dots(ii)
[latex]
Squaring and adding (i) and (ii), we get
[latex]
\begin{aligned}
& \frac{x^2}{a^2}+\frac{v^2}{a^2 \omega^2}=\sin ^2 \omega t+\cos ^2 \omega t \\
& \frac{x^2}{a^2}+\frac{v^2}{a^2 \omega^2}=1
\end{aligned}
\)
It is an equation of ellipse.
Hence, the graph between velocity and displacement is an ellipse.
Momentum of the particle \(=\mathrm{mv}\)
\(\therefore\) The nature of graph of the momentum and displacement is the same as that of velocity and displacement.

Hint

(c) Given,
Damping force \(\propto\) velocity

\(F \propto v \Rightarrow \quad F=k v\)
\(
k=\frac{F}{v} \Rightarrow[k]=\frac{\left[\mathrm{kgms}^{-2}\right]}{\left[\mathrm{ms}^{-1}\right]}=\mathrm{kgs}^{-1}
\)

Hint

(b) \(y=3 \sin \frac{\pi}{2}(50 t-x)\)
\(y=3 \sin \left(25 \pi t-\frac{\pi}{2} x\right)\) on comparing with the standard wave equation \(y=a \sin (\omega t-k x)\)
Wave velocity \(v=\frac{\omega}{k}=\frac{25 \pi}{\pi / 2}=50 \mathrm{~m} / \mathrm{sec}\).
The velocity of particle
\(
\begin{aligned}
& v_p=\frac{\partial y}{\partial t}=75 \pi \cos \left(25 \pi t-\frac{\pi}{2} x\right) \\
& v_{p \max }=75 \pi \\
& \text { then } \frac{v_{p_{\max }}}{v}=\frac{75 \pi}{50}=\frac{3 \pi}{2}
\end{aligned}
\)

Hint

(c) Only functions given in (A) & (C) represent SHM.
\(y=A \sin \omega t+B \cos \omega t \Rightarrow y=a \sin (\omega t+\phi)\). This is also a equation of S.H.M where \(\phi=\sqrt{A^2+B^2}\) \(\mathrm{A}=\mathrm{a} \cos \phi, \mathrm{B}=\mathrm{a} \sin \phi\) and \(\theta=\tan ^{-1} \frac{B}{A}\)

\(
\begin{aligned}
& \text { (a) } y=\sin w t-\cos w t \\
& =\sqrt{ } 2\left\{\frac{1}{\sqrt{2}} \sin w t-\frac{1}{\sqrt{2}} \cos w t\right\} \\
& =\sqrt{ } 2\left\{\cos 45^{\circ} \cdot \sin w t-\sin 45^{\circ} \cdot \cos w t\right\} \\
& =\sqrt{ } 2 \sin \left(w t-45^{\circ}\right)
\end{aligned}
\)
This is in the form of \(\mathrm{y}=\operatorname{Asin}(\mathrm{wt} \pm \emptyset)\)
So, this is the equation of SHM .
Period \(=\frac{2 \pi}{\omega}\)
(b) \(\sin ^3 w t\)
Use formula of \(\sin 3 x=3 \sin x-4 \sin ^3 x\)
So, \(\sin ^3 w t=\frac{1}{4}[3 \sin w t-\sin 3 w t]\)
Hence, you observed that this equation is a combination of two SHM. Hence, this is not SHM. But periodic motion.
Period = LCM of period \(\{\) sinwt, \(\sin 3 \mathrm{wt}\}\)
Period of sinwt \(=\frac{2 \pi}{\omega}\)
period of \(\sin 3 \omega t=\frac{2 \pi}{3 \omega}\)
so, period \(=\frac{2 \pi}{\omega}\)
(c) \(5 \cos \left(\frac{3 \pi}{4}-3 w t\right)\)
\(=5 \cos \left\{-\left(3 w t-\frac{3 \pi}{4}\right)\right\}\)
\(=5 \cos \left(3 \omega t-\frac{3 \pi}{4}\right)[\cos (-\emptyset)=\cos \emptyset]\)
Hence, this is SHM .
Period \(=\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}\)
(d) \(x=1+w t+w^2 t^2\)
At \(x \rightarrow \infty, t \rightarrow \infty\)
There is no repetition of values. hence, this is neither periodic nor SHM.

Hint

(d)
According to question,
\(\begin{array}{ll} & y_1=a \sin (\omega t+k x+0.57) \\ \text { and } & y_2=a \cos (\omega t+k x) \\ \text { or } & y_2=a \sin \left(\frac{\pi}{2}+\omega t+k x\right)\end{array}\)
As phase difference,
\(
\begin{aligned}
\Delta \phi & =\phi_2-\phi_1=\frac{\pi}{2}-0.57 \\
& =1.57-0.57=1 \mathrm{rad}
\end{aligned}
\)

Hint

We have potential energy \(V=x^2\)
\(
\begin{aligned}
& \Rightarrow F=-\frac{d v}{d x}=-\frac{d}{d x}\left(x^2\right)=-2 x \\
& \Rightarrow \quad \frac{m^2 x}{d t^2}=-2 x \\
& \Rightarrow\left(\frac{d^2}{d t^2}+\frac{2}{m} x\right)=0 \\
\end{aligned}
\)
\(
\Rightarrow \quad x=A \sin (\omega t+\phi)
\)
\(
\begin{aligned}
& \text { At } t=0, x_{\max } \Rightarrow x_{\max }=A(\sin \phi) \\
& \Rightarrow A=A \sin \phi \Rightarrow \phi=\frac{\pi}{2} \\
& \Rightarrow \quad x=A \cos \omega t
\end{aligned}
\)

The given velocity-position graph depicts that the motion of the particle is SHM.
In SHM, at \(t=0, \mathrm{v}=0\) and \(\mathrm{x}=\mathrm{x}_{\text {max }}\)
So, option (a) is correct.

Hint

(a) \(\mathrm{x}=\mathrm{a} \sin ^2 \omega \mathrm{t}=\frac{a}{2}(1-\cos 2 \omega \mathrm{t})\)
\(
\begin{aligned}
& \frac{d x}{d t}=\frac{a}{2} 2 \omega \sin 2 \omega t \\
& \Rightarrow \quad \frac{d^2 x}{d t^2}=\frac{4 \omega^2 a}{2} \cdot \cos 2 \omega t
\end{aligned}
\)
This represents an S. H. M. of frequency
\(
=\frac{\omega}{\pi}
\)

Hint

(b) Equation of SHM is given by \(x=A \sin (\omega t+\delta)\) \((\omega \mathrm{t}+\delta)\) is called phase.
When \(x=\frac{A}{2}\), then \(\sin (\omega t+\delta)=\frac{1}{2}\) \(\Rightarrow \omega \mathrm{t}+\delta=\frac{\pi}{6}\)
or \(\phi_1=\frac{\pi}{6}\)
For second particle,

\(
\begin{aligned}
& \phi_2=\pi-\frac{\pi}{6}=\frac{5 \pi}{6} \\
& \therefore \phi=\phi_2-\phi_1 \\
& =\frac{4 \pi}{6}=\frac{2 \pi}{3}
\end{aligned}
\)

Hint

(d)
Time period of spring pendulum,
\(
T=2 \pi \sqrt{\frac{M}{k}} \text {. }
\)
If mass is doubled then time period
\(
T^{\prime}=2 \pi \sqrt{\frac{2 M}{k}}=\sqrt{2} T
\)

Hint

(b)
As we know that, the condition for a body executing SHM is \(F=-k x\)
So,
\(
a=\frac{F}{m}=-\frac{k}{m} x
\)
or \(\quad a=-\omega^2 x\)
Acceleration \(\propto-\) (displacement)
\(
\begin{aligned}
& A \propto-y \\
& A=-\omega^2 y \\
& A=-\frac{k}{m} y \\
& A=-k y
\end{aligned}
\)
Here, \(y=x+a\)
\(\therefore \quad\) Acceleration \(=-k(x+a)\)

Hint

(d)
Concept Use the equation of motion of a body executing SHM.
i.e. \(\quad x=a \sin \omega t\)
As we know, the velocity of body executing SHM is given by
\(
\begin{aligned}
v & =\frac{d x}{d t}=a \omega \cos \omega t=a \omega \sqrt{1-\sin ^2 \omega t} \\
& =\omega \sqrt{a^2-x^2}
\end{aligned}
\)
Here, \(\quad x=\frac{a}{2}\)
\(
\begin{aligned}
\therefore \quad v & =\omega \sqrt{a^2-\frac{a^2}{4}}=\omega \sqrt{\frac{3 a^2}{4}} \\
& =\frac{2 \pi}{T} \frac{a \sqrt{3}}{2}=\frac{\pi a \sqrt{3}}{T}
\end{aligned}
\)

Hint

(d) Restoring force, \(f^{\prime}=-k x\)
where \(x\) is the extension produced in the spring.
Weight of the mass acting downward \(=\mathrm{Mg}\).
In equilibrium
\(
\mathrm{kx}=\mathrm{Mg} \quad \text { or } \quad \mathrm{x}=\frac{\mathrm{Mg}}{\mathrm{k}}
\)

Hint

(b) Maximum acceleration of a particle in the simple harmonic motion is directly proportional to the square of angular frequency i.e.,
\(
\text { i.e., a } \propto \mathrm{\omega}^2
\)
\(
\begin{aligned}
& \therefore \frac{a_1}{a_2}=\frac{\omega_1^2}{\omega_2^2}=\frac{(100)^2}{(1000)^2}=\frac{1}{100} \\
& \Rightarrow a_1: a_2=1: 10^2 .
\end{aligned}
\)

Hint

(d) We have \(x=\operatorname{asin}\left(\omega t+\frac{\pi}{6}\right)\)
\(\therefore\) Velocity, \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a} \omega \cos \left(\omega \mathrm{t}+\frac{\pi}{6}\right)\)
Maximum velocity \(=a \omega\)
According to question,
\(
\frac{a \omega}{2}=a \omega \cos \left(\omega t+\frac{\pi}{6}\right)
\)
or, \(\cos \left(\omega t+\frac{\pi}{6}\right)=\frac{1}{2}=\cos 60^{\circ}\) or \(\cos \frac{p}{3}\)
\(\Rightarrow \mathrm{wt}+\frac{\mathrm{p}}{6}=\frac{\mathrm{p}}{3}\)
\(w t=\frac{p}{3}-\frac{p}{6}\) or, \(w t=\frac{p}{6}\)
\(
\text { or, } \frac{2 \mathrm{p}}{\mathrm{T}} \cdot \mathrm{t}=\frac{\mathrm{p}}{6} \Rightarrow \mathrm{t}=\frac{\mathrm{T}}{12}
\)

Hint

(c) We have, \(U+K=E\)
where, \(U=\) potential energy, \(K=\) Kinetic energy, \(E=\) Total energy.
Also, we know that, in S.H.M., when potential energy is maximum, K.E. is zero and vice-versa.
\(
\therefore U_{\max }+0=E \Rightarrow U_{\max }=E
\)
Further,
\(
K . E .=\frac{1}{2} m \omega^2 a^2 \cos ^2 \omega t
\)
But by question, \(K . E .=K_0 \cos ^2 \omega t\)
\(
\therefore K_0=\frac{1}{2} m \omega^2 a^2
\)
Hence, total energy, \(E=\frac{1}{2} m \omega^2 a^2=K_0\)
\(
\therefore U_{\max }=K_0,  \& E=K_0 \text {. }
\)

Hint

(c)
Let displacement equation of particle executing SHM is
\(
x=a \sin \omega t
\)
As particle travels half of the amplitude from the equilibrium position, so
\(
x=\frac{a}{2}
\)
Therefore, \(\frac{a}{2}=a \sin \omega t\)
or \(\quad \sin \omega t=\frac{1}{2}=\sin \frac{\pi}{6}\)
or \(\omega t=\frac{\pi}{6}\) or \(t=\frac{\pi}{6 \omega}\)
or \(\quad t=\frac{\pi}{6\left(\frac{2 \pi}{T}\right)} \quad\left(\right.\) as, \(\left.\omega=\frac{2 \pi}{T}\right)\)
or \(t=\frac{T}{12}\)
Hence, the particle travels half of the amplitude from the equilibrium in \(\frac{T}{12} \mathrm{~s}\).
This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.

Hint

(b)
Let the minimum amplitude of SHM be \(a\).
Restoring force on spring
\(
F=k a
\)
Restoring force is balanced by weight \(\mathrm{mg}\) of block. For mass to execute simple harmonic motion of amplitude \(a\).
\(
\begin{aligned}
& \therefore k a=m g \text { or } a=\frac{m g}{k} \\
& \text { Here, } m=2 \mathrm{~kg}, k=200 \mathrm{~N} / \mathrm{m} \text {, } \\
& g=10 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore \quad a=\frac{2 \times 10}{200}=\frac{10}{100} \mathrm{~m} \\
& =\frac{10}{100} \times 100 \mathrm{~cm}=10 \mathrm{~cm} \\
&
\end{aligned}
\)
Hence, minimum amplitude of the motion should be \(10 \mathrm{~cm}\), so that the mass gets detached from the pan.

Hint

(d) Let \(y=A \sin \omega t\)
\(
v_{\text {inst }}=\frac{d y}{d t}=A \omega \cos \omega t=A \omega \sin (\omega t+\pi / 2)
\)
Acceleration \(=-A \omega^2 \sin \omega t\)
\(
=\mathrm{A} \omega^2 \sin (\pi+\omega t)
\)
\(
\therefore \phi=\frac{\pi}{2}=0.5 \pi
\)

Hint

(b)
Let block be displaced through \(x \mathrm{~m}\), then weight of displaced water or upthrust, (upwards) is given by Archimedes’ principle
\(
F_b=-\operatorname{Ax \rho g}
\)
where, \(A\) is the a
rea of cross-section of the block and pis its density. This must be equal to force (=ma) applied, where, \(m\) is the mass of the block and \(a\) is the acceleration.
\(
\therefore m a=-A \times \rho g \text { or } a=-\frac{A \rho g}{m} x=-\omega^2 x
\)
This is the equation of simple harmonic motion. Time period of oscillation,
\(
T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{A \rho g}} \Rightarrow T \propto \frac{1}{\sqrt{A}}
\)

Hint

(b) The potential energy of a spring \(=\frac{1}{2} k x^2\) \(U=\frac{1}{2} k \cdot(2)^2=4 \times \frac{1}{2} k\) For \(\mathrm{x}=8 \mathrm{~cm}\), Energy stored \(=\frac{1}{2} k \cdot(8)^2=64 \times \frac{1}{2} k\) \(=64 \times \frac{U}{4}=16 U\)

Hint

(d)
Maximum speed of a particle executing SHM is given by,
\(
v_{\max }=a \omega=a(2 \pi n) \Rightarrow n=\frac{v_{\max }}{2 \pi a}
\)
where, \(a=\) amplitude of oscillation
\(n=\) frequency of oscillation
Here, \(v_{\max }=31.4 \mathrm{~cm} / \mathrm{s}, a=5 \mathrm{~cm}\)
Substituting, the given values, we have
\(
n=\frac{31.4}{2 \times 3.14 \times 5}=1 \mathrm{~Hz}
\)

Hint

(d)
In simple harmonic motion, the displacement equation is, \(x=a \sin \omega t\) where, \(a\) is the amplitude of the motion.
Velocity, \(\quad v=\frac{d x}{d t}=a \omega \cos \omega t\)
\(v=a \omega \sqrt{1-\sin ^2 \omega t}\)
\(v=\omega \sqrt{a^2-x^2} \dots(i)\)
\(
\begin{gathered}
\text { Acceleration, } \alpha=\frac{d v}{d t}=\frac{d}{d t}(a \omega \cos \omega t) \\
\alpha=-a \omega^2 \sin \omega t \\
\alpha=-\omega^2 x
\end{gathered}
\)
When \(x=0, v=a \omega=v_{m x}\)
\(
\alpha=0=\alpha_{\text {min }}
\)
When \(x=a, v=0=v_{\mathrm{mn}}\)
\(
\alpha=-\omega^2 a=\alpha_{\max }
\)
Hence, it is clear that when \(v\) is maximum, then \(\alpha\) is minimum (i.e. zero) or vice-versa.

Hint

(d)
Let us consider two springs of spring constants \(k_1\) and \(k_2\) joined in series as shown in the figure.
Under a force \(F\), they will stretch by \(y_1\) and \(y_2\).

So, \(\quad y=y_1+y_2\)
or \(\frac{F}{k}=\frac{F_1}{k_1}+\frac{F_2}{k_2}\)
But as springs are massless, so force on them must be the same i.e. \(F_1=F_2=F\).
So, \(\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}\) or \(k=\frac{k_1 k_2}{k_1+k_2}\)

Hint

(d) \(P \cdot E .=\frac{1}{2} k x^2=E\)
At half way
\(
P . E .=\frac{1}{2} k\left(\frac{x}{2}\right)^2=\frac{\frac{1}{2} k x^2}{4}=\frac{E}{4}
\)

Hint

Potential energy is given by \(U=\frac{1}{2} k x^2\)
The corresponding graph is shown in figure.

At equilibrium position \((x=0)[latex], potential energy is minimum. At extreme positions [latex]x_1\) and \(x_2\), its potential energies are
\(
\begin{aligned}
U_1 & =\frac{1}{2} k x_1^2 \\
\text { and } \quad U_2 & =\frac{1}{2} k x_2^2
\end{aligned}
\)

In the above graph, the dotted line (curve) is shown for kinetic energy. This graph shows that kinetic energy is maximum at mean position and zero at extreme positions \(x_1\) and \(x_2\).

Note:

Potential energy \(\propto(\text { Amplitude })^2\)
\(\propto\) (Displacement \()^2\)
P.E.is maximum at maximum distance.
P.E. is zero at equilibrium point.
P.E. curve is parabolic.

Hint

(c) \(T=2 \pi \sqrt{\frac{m}{k}}\)
When a spring is cut into \(\mathrm{n}\) parts Spring constant for each part \(=n k\) Here, \(n=4\)
\(
T_1=2 \pi \sqrt{\frac{m}{4 k}}=\frac{T}{2}
\)

Hint

(b) The resonance wave becomes very sharp when damping force is small.

Forced vibration
1. Such vibration where a body or a system is made to vibrate under the action of a strong impressed periodic force with a frequency equal to that of the impressed force, which is different from the natural frequency of the system, is called forced vibration.
2. We know that the amplitude of forced vibration is defined by the form, \(a=\frac{f_o}{2 k \omega}\), where, \(f_o=\frac{F_o}{m}\) ( \(F_o\) is the driving force at time zero and \(m\) is the mass of the oscillator), \(k\) is the damping coefficient, and \(\omega\) is the frequency.

Amplitudes at resonance
1. As we know, the amplitude of resonance is inversely proportional to the damping coefficient \(k\), amplitude, \(a=\frac{f_o}{2 k \omega}\).
2. Resonance will fast when amplitude will maximum. The amplitude will be maximum when the damping coefficient will minimum.
3. So, at \(k=0, a=\infty\) i.e. if we decrease the damping force, the resonance wave becomes very fast.

On applying forced vibrations, the resonance wave becomes very fast when the damping force is small.

Hint

(b) P.E. of particle executing S.H.M. \(=\frac{1}{2} m \omega^2 x^2\)
At \(x=a\), P.E. is maximum i.e. \(=\frac{1}{2} m \omega^2 a^2\) K.E. \(=\frac{1}{2} m \omega^2\left(a^2-x^2\right)\)
At \(x=0\), K.E. is maximum. Hence, displacement from position of maximum Kinetic energy \(= \pm a\).

Hint

\(
\text { (d) The time period of spring mass system, }
\)
\(
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}
\)
\(
\therefore \quad \mathrm{t}_1=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_1}} \dots(i)
\)
\(
\mathrm{t}_2=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_2}} \dots(ii)
\)
When springs are connected in parallel then \(\mathrm{K}_{\mathrm{eff}}=\mathrm{k}_1+\mathrm{k}_2\)
So, \(\mathrm{t}_0=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{\mathrm{eff}}}} \Rightarrow 2 \pi \sqrt{\frac{\mathrm{m}}{\left(\mathrm{k}_{1+\mathrm{k}_2}\right)}}\dots(iii)\)
from (i), \(\frac{1}{\mathrm{t}_1^2}=\frac{1}{4 \pi^2} \times \frac{\mathrm{k}_1}{\mathrm{~m}}\)
from(ii), \(\frac{1}{t_2^2}=\frac{1}{4 \pi^2} \times \frac{k_2}{m}\)
from (iii), \(\frac{1}{\mathrm{t}_0^2}=\frac{1}{4 \pi^2} \times \frac{\left(\mathrm{k}_1+\mathrm{k}_2\right)}{\mathrm{m}}\)
from above expressions,
\(
\frac{1}{t_0^2}=\frac{1}{t_1^2}+\frac{1}{t_2^2}
\)

Notes:

\(
\begin{aligned}
& t=2 \pi \sqrt{\frac{m}{k}} \\
& \Rightarrow k=\text { Const } . t^{-2}
\end{aligned}
\)
Here the springs are joined in parallel. So
\(
k_0=k_1+k_2
\)
where \(\mathrm{k}_0\) is resultant force constant
\(\therefore\) Const. \(t_0^{-2}=\) Const. \(t_1^{-2}+\) Const. \(t_2^{-2}\)
or, \(t_0^{-2}=t_1^{-2}+t_2^{-2}\)

Hint

In case of damped vibration, amplitude at any instant \(t\) is given by
\(a=a_0 e^{-b t}\)
where, \(a_0=\) initial amplitude
\(b=\) damping constant
Case-1: \(t=100 T\) and \(a=\frac{a_0}{3}\)
\(
\begin{aligned}
& \therefore \quad \frac{a_0}{3}=a_0 e^{-b(100 T)} \\
& \Rightarrow \quad e^{-100 \text { bT }}=\frac{1}{3} \\
& \text { Case II: } t=200 T \\
& a=a_0 e^{-b t}=a_0 e^{-b(200 T)} \\
& =a_0\left(e^{-100 b T}\right)^2=a_0 \times\left(\frac{1}{3}\right)^2=\frac{a_0}{9} \\
&
\end{aligned}
\)
Thus, after 200 oscillations, amplitude will become \(\frac{1}{9}\) times.

Hint

In SHM, the total energy
\(=\) potential energy + kinetic energy
or
\(
\begin{aligned}
E & =U+K \\
& =\frac{1}{2} m \omega^2 x^2+\frac{1}{2} m \omega^2\left(a^2-x^2\right) \\
& =\frac{1}{2} m \omega^2 a^2=\frac{1}{2} k a^2
\end{aligned}
\)
where,\(k=\) force constant \(=m \omega^2\)
Thus, total energy depends on \(k\) and \(a\).

Hint

(b) We know that the time period (T)
\(
=2 \pi \sqrt{\frac{l}{g^{\prime}}} \propto \frac{1}{\sqrt{g^{\prime}}} \quad \mathrm{g}^{\prime}=\mathrm{g}_{\text {effective }}
\)
Since the negatively charged bob is attracted by the positively charged plate, therefore acceleration due to gravity will increase and time period will decrease.

Hint

Two simple harmonic motions can be written as
\(
\begin{gathered}
x=a \sin (\omega t+\delta) \dots(i)\\
\text { and } \quad y=a \sin \left(\omega t+\delta+\frac{\pi}{2}\right)
\end{gathered}
\)
or \(\quad y=a \cos (\omega t+\delta) \dots(ii)\)
Squaring and adding Eqs. (i) and (ii), we obtain
\(
x^2+y^2=a^2\left[\sin ^2(\omega t+\delta)+\cos ^2(\omega t+\delta)\right]
\)
or \(x^2+y^2=a^2\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right)\)
This is the equation of a circle.
\(
\begin{aligned}
& \text { At }(\omega t+\delta)=0 ; x=0, y=a \\
& \operatorname{At}(\omega t+\delta)=\frac{\pi}{2} ; x=a, y=0 \\
& \operatorname{At}(\omega t+\delta)=\pi ; x=0, y=-a \\
& \operatorname{At}(\omega t+\delta)=\frac{3 \pi}{2} ; x=-a, y=0 \\
& \operatorname{At}(\omega t+\delta)=2 \pi ; x=0, y=a
\end{aligned}
\)
Thus, it is obvious that the motion of the particle is traversed in the clockwise direction.

Hint

If \(l\) is the length of pendulum and \(\theta\) the angular amplitude, then height

\(
\begin{aligned}
h & =A B-A C \\
& =l-l \cos \theta \\
& =l(1-\cos \theta) \ldots(i) \\
\end{aligned}
\)
At the point \(P\) (maximum displacement position i.e. extreme position), potential energy is maximum and kinetic energy is zero. At the point \(B\) (mean or equilibrium position) potential energy is minimum and kinetic energy is maximum, so from the principle of conservation of energy.
\(
(P E+K E) \text { at } P=(K E+P E) \text { at } B
\)
or \(m g h+0=\frac{1}{2} m v^2+0\)
or \(\quad v=\sqrt{2 g h} \dots(ii)\)
Substituting the value of \(h\) from Eq. (i) into Eq. (ii), we get
\(
v=\sqrt{2 g l(1-\cos \theta)}
\)

Hint

(c)
\(f_A=\frac{1}{2 \pi} \sqrt{\frac{g}{L_A}}\) and \(f_B=\frac{f_A}{2}=\frac{1}{2 \pi} \sqrt{\frac{g}{L_B}}\)
\(
\begin{aligned}
& \therefore \frac{f_A}{f_{A / 2}}=\frac{1}{2 \pi} \sqrt{\frac{g}{L_A}} \times 2 \pi \sqrt{\frac{L_B}{g}} \\
& \Rightarrow 2=\sqrt{\frac{L_B}{L_A}} \Rightarrow 4=\frac{L_B}{L_A},
\end{aligned}
\)
regardless of mass.

Hint

(d) \(T=2 \pi \sqrt{\frac{\ell}{g}} \quad T \propto \sqrt{\ell}\)
If \(\ell\) is increased by 4 times, time period will increase by two times.

Explanation:

Time period of simple pendulum
\(
T=2 \pi \sqrt{\frac{l}{g}}
\)
where, \(l\) = length of pendulum
\(g=\) acceleration due to gravity
i.e. \(T \propto \sqrt{l}\)
Hence, \(\quad \frac{T_2}{T_1}=\sqrt{\frac{l_2}{l_1}} \dots(i)\)
Given, \(l_2=4 l_1, T_1=2 \mathrm{~s}\)
Substituting the values in Eq. (i), we get
\(
T_2=\sqrt{\frac{4 l_1}{l_1}} \times 2=2 \times 2=4 \mathrm{~s}
\)

Hint

(b) We know that if the frequency of an external forced oscillation is equal to the natural frequency of the body, then the amplitude of the forced oscillation of the body becomes very large. This phenomenon is known as resonant vibration. Therefore, resonance is an example of forced vibration.

Hint

(a) Amplitude \((\mathrm{A})=0.01 \mathrm{~m}\), Frequency \(=60 \mathrm{~Hz}\)
Maximum acceleration
\(
\begin{aligned}
& =A \omega^2=0.01 \times(2 \pi n)^2 \\
& =0.01 \times 4 \pi^2 \times 60 \times 60=144 \pi^2 \mathrm{~m} / \mathrm{sec}^2
\end{aligned}
\)

Hint

(c) \(n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\)
\(
n^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{k}{4 m}}=\frac{1}{2} \times \frac{1}{2 \pi} \sqrt{\frac{k}{m}}
\)
On putting the value of \(\mathrm{n}\) we get \(n^{\prime}=\frac{n}{2}\)

Hint

(c) Let the pendulums be in phase after \(t \mathrm{sec}\) of start. Within this time, if the bigger pendulum executes \(n\) oscillations, the smaller one will have executed \((n+1)\) oscillations.
Now, the time of \(n\) oscillation \(=2 \pi \sqrt{\frac{20}{g}} \times n\) \(\&\) the time of \((n+1)\) oscillation
\(
=2 \pi \sqrt{\frac{5}{g}} \times(n+1)
\)
To be in phase
\(
2 \pi \sqrt{\frac{20}{g}} \times n=2 \pi \sqrt{\frac{5}{g}} \times(n+1)
\)
or, \(\quad 2 n=n+1\)
or, \(n=1\)
Hence, the no. of oscillations executed by shorter pendulum \(=n+1=1+1=2\)

Hint

(a) Equation of two simple harmonic motions
\(
\begin{aligned}
& y=A \sin (\omega t+\phi) \dots(1)\\
& x=A \sin \left(\omega t+\phi+\frac{\pi}{2}\right) \\
& \Rightarrow x=A \cos (\omega t+\phi) \dots(2)
\end{aligned}
\)
On squaring and adding equations (1) and (2), \(x^2+y^2=A^2\)
This is an equation of a circle. Hence, the resulting motion will be a circular motion.

Hint

(a) Time period of the simple pendulum
\(T=2 \pi \sqrt{\left(\frac{l}{g}\right)} \propto \sqrt{l}\) where \(l\) is effective length.
[i.e distance between centre of suspension and centre of gravity of bob]
Initially, centre of gravity is at the centre of the sphere. When water leaks the centre of gravity goes down until it is half filled; then it begins to go up and finally, it again goes at the centre. That is effective length first increases and then decreases. As \(T \propto \sqrt{l}\), so time period first increases and then decreases.

Hint

(c) We know that \(T=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\)
\(
\frac{\Delta T}{T} \times 100=\frac{1}{2} \frac{\Delta l}{l} \times 100
\)
If length is increased by \(2 \%\), time period increases by \(1 \%\).

Notes:

(i) If \(g\) is constant and length varies by \(n \%\). Then \(\%\) change in time period \(\frac{\Delta T}{T} \times 100=\frac{n}{2} \%\)
(ii) If \(l\) is constant and \(g\) varies by \(n \%\). Then \(\%\) change in time period \(\frac{\Delta T}{T} \times 100=\frac{n}{2} \%\) Valid only for percentage less than \(10 \%\).

Hint

The velocity of the particle executing SHM at any instant is defined as the time rate of change of its displacement at that instant.
Let the displacement of the particle at an instant \(t\) is given by
\(
\begin{gathered}
x=a \sin \omega t \\
\therefore \text { Velocity  v }=\frac{d x}{d t}=\frac{d(a \sin \omega t)}{d t} \\
=a \omega \cos \omega t=a \omega \sqrt{\left(1-\sin ^2 \omega t\right)} \\
=a \omega \sqrt{\left(1-\frac{x^2}{a^2}\right)}=\omega \sqrt{\left(a^2-x^2\right)}
\end{gathered}
\)
At mean position, \(x=0\)
\(
\therefore \quad v_{\max }=\omega a
\)
According to problem, \(v=\frac{v_{\max }}{2}=\frac{a \omega}{2}\)
But \(\quad v=\omega \sqrt{a^2-x^2}\)
\(
\therefore \quad \frac{a \omega}{2}=\omega \sqrt{a^2-x^2} \text { or } x=\frac{\sqrt{3}}{2} a
\)

Hint

(d) Total energy of particle executing S.H.M. of amplitude (a).
\(
\mathrm{E}=\frac{1}{2} m \omega^2 a^2
\)
K.E. of the particle
\(
\begin{aligned}
& =\frac{1}{2} m \omega^2\left(a^2-\frac{a^2}{4}\right)\left(\text { when } x=\frac{a}{2}\right) \\
& =\frac{1}{2} m \omega^2 \times \frac{3}{4} a^2=\frac{1}{2} \times \frac{3}{4} m \omega^2 a^2 \\
& \text { Clearly, } \frac{\mathrm{KE}}{\text { Total Energy }}=\frac{3}{4}
\end{aligned}
\)

Hint

Given,
\(
\begin{aligned}
& \mathrm{A}=0.01 \mathrm{~m} \\
& \mathrm{k}=2 \times 10^6 \mathrm{~N} / \mathrm{m} \\
& \mathrm{E}_{\mathrm{T}}=160 \mathrm{~J}
\end{aligned}
\)
The total mechanical energy,
\(
\mathrm{E}_{\mathrm{T}}=(\mathrm{K} . \mathrm{E})_{\max }+(\mathrm{P} . \mathrm{E})_{\min }=160 \mathrm{~J} \ldots \ldots(1)
\)
we know that, maximum kinetic energy,
\(
\begin{aligned}
& (\mathrm{K} . \mathrm{E})_{\max }=\frac{1}{2} \mathrm{kA}^2 = \frac{1}{2} \times 2 \times 10^6 \times 10^{-6} \\
& (\text { K.E })_{\max }=100 \mathrm{~J}
\end{aligned}
\)
from equation (1),
\(
(\text { P.E })_{\min }=160-100=60 \mathrm{~J}
\)
The minimum kinetic energy is zero so, the maximum potential energy is \(160 \mathrm{~J}\)

Hint

Potential energy of the body executing SHM is given by
\(
U=\frac{1}{2} m \omega^2 x^2
\)
where symbols have their usual meaning.
Total energy of the body executing SHM is
\(
E=\frac{1}{2} m \omega^2 a^2
\)
According to problem,
\(
\begin{gathered}
U=\frac{1}{4} E \\
\therefore \quad \frac{1}{2} m \omega^2 x^2=\frac{1}{4} \times \frac{1}{2} m \omega^2 a^2 \\
\text { or } \quad x^2=\frac{a^2}{4} \text { or } x=\frac{a}{2}
\end{gathered}
\)

Hint

In transverse wave motion individual particles of the medium execute simple harmonic motion about their mean position in a direction perpendicular to the direction of propagation of wave motion. Wave moving through a string fixed at both ends executes SHM.

Hint

(c) As phase difference \(=\frac{\pi}{4}\), the resultant path of particle is an ellipse.

Explanation:

Hence \(\sin \omega t=\frac{x}{2}\)
\(
\therefore \quad \cos \omega \mathrm{t}=\frac{\sqrt{4-\mathrm{x}^2}}{2}
\)
\(\therefore \quad y=2 \sin \omega t \cos \frac{\pi}{4}+2 \cos \omega t \sin \frac{\pi}{4}\)
or \(\quad \mathrm{y}=\frac{2}{\sqrt{2}} \sin \omega \mathrm{t}+\frac{2}{\sqrt{2}} \cos \omega \mathrm{t}\)
\(
y=\sqrt{2} \times \frac{x}{2}+\sqrt{2} \frac{\sqrt{4-x^2}}{2}
\)
or \(y=\frac{x}{\sqrt{2}}+\frac{\sqrt{4-x^2}}{\sqrt{2}}\)
or \(\sqrt{2} \mathrm{y}-\mathrm{x}=\sqrt{4-\mathrm{x}^2}\)
or \(2 y^2+x^2-2 \sqrt{2} x y=4-x^2\)
or \(2 y^2+2 x^2-2 \sqrt{2} x y=4\)
or \(x^2+y^2-\sqrt{2} x y=2\)
This is an equation of an ellipse.

Hint

Frequency \(=\frac{1}{\text { Time period }}\)
Beat frequency \(=\frac{1}{\Delta t}=f_1-f_2=\frac{1}{T_1}-\frac{1}{T_2}\)
\(\Delta t\) is the time period of the resultant wave
\(\Rightarrow\) Given:- Time period of wave \(1=T_1=3 \mathrm{~sec}\)
Time period of wave \(2=T_2=4 \mathrm{~sec}\)
We have to find the Time period of resultant ware \(\Delta t=\)?
\(
\begin{aligned}
\Rightarrow \quad \frac{1}{\Delta t} & =\frac{1}{T_1}-\frac{1}{T_2}=\frac{1}{3}-\frac{1}{4} \\
\frac{1}{\Delta t} & =\frac{1}{12} \\
\Delta t & =\text { Time period of resultant ware }=12 \mathrm{~sec}
\end{aligned}
\)
option \(C\) is correct answer.

Hint

(a) For S.H.M., \(x=A \sin \left(\frac{2 \pi}{T} t\right)\)
When \(x=A, A=A \sin \left(\frac{2 \pi}{T} . t\right)\)
\(
\begin{aligned}
& \therefore \sin \left(\frac{2 \pi}{T} \cdot t\right)=1 \\
& \Rightarrow \sin \left(\frac{2 \pi}{T} \cdot t\right)=\sin \left(\frac{\pi}{2}\right) \Rightarrow t=(T / 4)
\end{aligned}
\)
When \(x=\frac{A}{2}, \frac{A}{2}=A \sin \left(\frac{2 \pi}{T} \cdot t\right)\) or, \(\sin \frac{\pi}{6}=\sin \left(\frac{2 \pi}{T} t\right)\) or \(t=(T / 12)\)
Now, time taken to travel from \(x=A\) to \(x=A / 2\) is \((T / 4-T / 12)=T / 6\)

Hint

Time period of a body executing SHM is given by
\(
T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{x}{a}} \dots(i)
\)
where, \(x\) is displacement of the particle and \(a\) is acceleration of the particle.
From Eq. (i)
\(
\omega=\sqrt{\frac{a}{x}} \text { or } \omega^2=\frac{a}{x}
\)
Here, \(a=2.0 \mathrm{~m} / \mathrm{s}^2\)
\(
\begin{aligned}
x & =0.02 \mathrm{~m} \\
\therefore \quad \omega^2 & =\frac{2.0}{0.02}
\end{aligned}
\)
or \(\quad \omega^2=100\)
or \(\quad \omega=10 \mathrm{~rad} / \mathrm{s}\)

Note:

\(\omega^2=\frac{\text { acceleration }}{\text { displacement }}=\frac{2.0}{0.02}\)
\(
\omega^2=100 \text { or } \omega=10 \mathrm{~rad} / \mathrm{s}
\)

Hint

Velocity of the particle executing SHM at any instant is defined as the time rate of change of its displacement at that instant. It is given by
\(
v=\omega \sqrt{\left(a^2-x^2\right)}
\)
where, \(x\) is displacement of the particle. is acceleration and \(\omega\) is angular frequency.
Case I: \(10=\omega \sqrt{a^2-16} \dots(i)\)
Case II: \(\quad 8=\omega \sqrt{a^2-25} \dots(ii)\)
Dividing Eq. (ii) by Eq. (i), we get
\(
\frac{5}{4}=\frac{\sqrt{a^2-16}}{\sqrt{a^2-25}} \text { or } \frac{25}{16}=\frac{a^2-16}{a^2-25}
\)
or \(16 a^2-256=25 a^2-625\)
or \(\quad a^2=\frac{369}{9}\)
\(
\text { Putting value of } a^2 \text { in Eq. (i), we get }
\)
\(
\begin{aligned}
& 10=\omega \sqrt{\left(\frac{369}{9}-16\right)} \\
& \text { or } \omega=\frac{10 \times 3}{15}=2 \mathrm{rad} / \mathrm{s} \\
& \therefore \text { Time period } T=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi \sec \\
&
\end{aligned}
\)

Hint

Consider a particle of mass \(m\), executing linear SHM with amplitude \(a\) and constant angular frequency \(\omega\). Supposet second af ter starting from the mean position, the displacement of the particle is \(x\), which is given by
\(
x=a \sin \omega t
\)
So, potential energy of particle is
\(
U=\frac{1}{2} m \omega^2 x^2 \dots(i)
\)
and kinetic energy of particle is
\(
T=\frac{1}{2} m \omega^2\left(a^2-x^2\right) \dots(ii)
\)
From Eqs. (i)and (ii) \(\frac{T}{U}=\frac{a^2-x^2}{x^2}\)

Hint

Apply the vector formula to determine the resultant acceleration of the both.

The bob is now under the combined action of two accelerations, \(g\) vertically downwards and \(a\) along the horizontal.
\(\therefore\) Resultant acceleration
\(
g^{\prime}=\sqrt{g^2+a^2}
\)

Hint

Let the SHM’s be
\(
x=a \sin \omega t \dots(i)
\)
and \(y=b \sin (\omega t+\pi)\)
or \(y=-b \sin \omega t \dots(ii)\)
From Eqs. (i)and (ii)
\(
\begin{gathered}
\frac{x}{a}=\sin \omega t \text { and }-\frac{y}{b}=\sin \omega t \\
\therefore \quad \frac{x}{a}=-\frac{y}{b} \text { or } y=-\frac{b}{a} x
\end{gathered}
\)
It is an equation of a straight line.

Hint

(d) The effective spring constant of two springs in series; \(K=\frac{k_1 k_2}{k_1+k_2}\).
Time period,
\(
T=2 \pi \sqrt{\frac{m}{K}}=2 \pi \sqrt{\frac{m\left(k_1+k_2\right)}{k_1 k_2}}
\)

Hint

(c) At maximum energy of the particle, velocity resonance takes place, which occurs when frequency of external periodic force is equal to natural frequency of undamped vibrations, i.e. \(\omega_2=\omega_0\).
Further, amplitude resonance takes place at a frequency of external force which is less than the frequency of undamped natural vibrations, i.e., \(\omega_1 \neq \omega_0\).

Hint

If a particle executing simple harmonic motion, has a displacement \(x\) from its equilibrium position, at an instant the magnitude of the restoring force \(F\) acting on the particle at that instant is given by
\(
F=-k x
\)
where, \(k\) is known as force constant. Hence, in given options, option (a) is correct. Here, \(k=A K\)

Hint

\(
\begin{aligned}
& f_o=f_{\text {open pipe }}=\frac{v}{2 l} \\
& f_c=f_{\text {closed pipe }}=\frac{v}{4 l} \\
& \frac{f_0}{f_c}=\frac{v}{2 l} \times \frac{4 l}{v} \\
& f_0: f_c=2: 1
\end{aligned}
\)

Hint

Position of particle as function of time \(x=A \sin \omega t\)
\(
\begin{aligned}
&\begin{aligned}
& T=8 \mathrm{~s} \\
& \omega=\frac{2 \pi}{T}=\frac{\pi}{4} \frac{\mathrm{rad}}{\mathrm{s}}
\end{aligned}\\
&\begin{aligned}
& a=-\omega^2 A \\
& a=-\left[\frac{\pi}{4}\right]^2 \times 1 \\
& a=-\frac{\pi^2}{16} \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\end{aligned}
\)