Sample Exercises-I

Hint

(a) It is not a periodic motion. Though the motion of a swimmer is to and fro but will not have a definite period.
(b) Since a freely suspended magnet if once displaced from N-S direction and released, it oscillates about this position, it is a periodic motion.
(c) The rotating motion of a hydrogen molecule about its centre of mass is periodic.
(d) Motion of an arrow released from a bow is non-periodic.

Hint

(b) and (c): SHM; (a) and (d) represent periodic but not SHM [A polyatomic molecule has a number of natural frequencies; so in general, its vibration is a superposition of SHM’s of a number of different frequencies. This superposition is periodic but not SHM].

Hint

(b) and (d) are periodic, each with a period of \(2 \mathrm{~s}\); (a) and (c) are not periodic. [Note in (c), repetition of merely one position is not enough for motion to be periodic; the entire motion during one period must be repeated successively].

Hint

(a) Simple harmonic, \(T=(2 \pi / \omega)\); (b) periodic, \(T=(2 \pi / \omega)\) but not simple harmonic;
(c) simple harmonic, \(T=(\pi / \omega)\); (d) periodic, \(T=(2 \pi / \omega)\) but not simple harmonic;

The function will represent a periodic motion, if it is identically repeated after a fixed interval of time and will represent S.H.M if it can be written uniquely in the form of a cos
\(\left(\frac{2 \pi t}{T}+\phi\right)\) or a \(\sin \left(\frac{2 \pi t}{T}+\phi\right)\), where \(T\) is the time period.
\(
\text { (a) } \begin{aligned}
\sin \omega t-\cos \omega t & =\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin \omega t-\frac{1}{\sqrt{2}} \cos \omega t\right] \\
& =\sqrt{2}\left[\sin \omega t \cos \frac{\pi}{4}-\cos \omega t \sin \frac{\pi}{4}\right] \\
& =\sqrt{2} \sin \left(\omega t-\frac{\pi}{4}\right)
\end{aligned}
\)
It is a S.H.M. and its period is \(2 \pi / \omega\)
(b) \(\sin ^3 \omega t\)
\(
=\frac{1}{3}[3 \sin \omega t-\sin 3 \omega t]
\)
Here each term \(\sin \omega t\) and \(\sin 3 \omega t\) individually represents S.H.M. But (ii) which is the outcome of the superposition of two SHMs will only be periodic but not SHMs. Its time period is \(2 \pi / \omega\).
(c) \(3 \cos \left(\frac{\pi}{4}-2 \omega t\right)=3 \cos \left(2 \omega t-\frac{\pi}{4}\right) . \quad[\because \cos (-\theta)=\cos \theta]\)
Clearly it represents SHM and its time period is \(2 \pi / 2 \omega\).
(d) \(\cos \omega t+\cos 3 \omega t+\cos 5 \omega t\). It represents the periodic but not S.H.M. Its time period is \(2 \pi / \omega\)

Hint

\(
\text { (c) non-periodic; (d) non-periodic (physically not acceptable as the function } \rightarrow \infty \text { as } t \rightarrow \infty \text {. }
\)

(a) \(3 \cos \left(\frac{\pi}{4}-2 \omega t\right)=3 \cos \left(2 \omega t-\frac{\pi}{4}\right) . \quad[\because \cos (-\theta)=\cos \theta]\)
Clearly it represents SHM and its time period is \(2 \pi / 2 \omega\).
(b) \(\cos \omega t+\cos 3 \omega t+\cos 5 \omega t\). It represents the periodic but not S.H.M. Its time period is \(2 \pi / \omega\)
(c) \(\mathrm{e}^{-\mathrm{w}^2 t^2}\). It is an exponential function which never repeats itself. Therefore it represents non-periodic motion.
(d) \(1+w t+w^2 t^2\) also represents non periodic motion.

Hint

In the fig. (given below), the points A and B, \(10 \mathrm{~cm}\) apart, are the extreme positions of the particle in SHM, and the point \(\mathrm{O}\) is the mean position. The direction from \(\mathrm{A}\) to \(\mathrm{B}\) is positive, as indicated.

(a) At the end A, i.e., extreme position, velocity is zero, acceleration and force are directed towards \(\mathrm{O}\) and are positive.
(b) At the end B, i.e., second extreme position, velocity is zero whereas the acceleration and force are directed towards the point \(O\) and are negative.
(c) At the mid point \(\mathrm{O}\), while going towards \(\mathrm{A}\), velocity is negative and maximum. The acceleration and force both are zero.
(d) At \(2 \mathrm{~cm}\) away from \(B\), that is, at \(C\) and going towards \(A\) : \(v\) is negative; acceleration and \(F\), being directed towards 0 , are also negative.
(e) At \(3 \mathrm{~cm}\) away from \(A\), that is, at \(D\) and going towards \(B\) : \(V\) is positive; acceleration and \(F\), being directed towards 0 , are also positive.
(f) At a distance of \(4 \mathrm{~cm}\) away from \(A\) and going towards \(A\), velocity is directed along BA, therefore, it is positive. Since acceleration and force are directed towards \(O B\), both of them are positive.

Hint

\(
\text { (c) represents a simple harmonic motion. }
\)

This is because acceleration is proportional and opposite to displacement (x).

Hint

The given displacement function is
\(
\begin{aligned}
& x(t)=A \cos (\omega t+\phi) \dots(i) \\
& \text { At } \quad t=0, \quad x(0)=1 \mathrm{~cm} \text {. Also, } \omega=\pi \mathrm{s}^{-1} \\
& \therefore \quad 1=A \cos (\pi \times 0+\phi) \\
& \Rightarrow \quad A \cos \phi=1 \dots(ii) \\
&
\end{aligned}
\)
Also, differentiating eqn. (i) w.r.t. ‘ \(t\) ‘.
\(
\begin{aligned}
& v=\frac{d}{d t} x(t)=-A \omega \sin (\omega t+\phi) \dots(iii) \\
& \text { Now at } \quad t=0, \quad v=\omega \\
& \therefore \text { from eqn. (iii), } \quad \omega=-A \omega \sin (\pi \times 0+\phi) \\
& \text { or } \quad A \sin \phi=-1 \dots(iv) \\
&
\end{aligned}
\)
Squaring and adding eqns. (ii) and (iv).
\(
A^2 \cos ^2 \phi+A^2 \sin ^2 \phi=1^2+1^2 \text { or } A=\sqrt{2} \mathrm{~cm}
\)
Dividing eqns. (ii) and (iv),
\(
\frac{A \sin \phi}{A \cos \phi}=\frac{-1}{1} \quad \therefore \quad \tan \phi=-1 \Rightarrow \phi=\frac{3 \pi}{4}
\)
If instead we use the sine function, i.e.,
\(
x=B \sin (\omega t+\alpha), \text { then } v=\frac{d}{d t} B \omega \cos (\omega t+\alpha)
\)
\(\therefore\) At \(\quad t=0\), using \(x=1\) and \(v=\omega\), we get \(1=B \sin (\omega \times 0+\alpha)\) or \(\quad B \sin \alpha=1 \dots(v)\)
and \(\quad \omega=B \omega \cos (\omega \times 0+\alpha)\) or \(B \cos \alpha=1 \dots(vi) \)
Dividing (v) by (vi),
\(
\tan \alpha=1 \quad \text { or } \alpha=\frac{\pi}{4} \quad \text { or } \quad \frac{5 \pi}{4}
\)
Squaring \((v)\) and \((v i)\), we get
\(
\begin{aligned}
B^2 \sin ^2 \alpha+B^2 \cos ^2 \alpha & =1^2+1^2 \\
\Rightarrow \quad B & =\sqrt{2} \mathrm{~cm} .
\end{aligned}
\)

Hint

\(
\begin{aligned}
M & =50 \mathrm{~kg}, \quad y=20 \mathrm{~cm}=0.2 \mathrm{~m}, T=0.60 \mathrm{~s} \\
F & =k y \quad \text { or } \quad M g=k y \quad \text { or } \quad k=\frac{M g}{0.2}=\frac{50 \times 9.8}{0.2} \mathrm{Nm}^{-1} \\
\mathrm{~K} & =2450 \mathrm{Nm}^{-1}
\end{aligned}
\)
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{m}{k}} \\
& T^2=4 \pi^2 \frac{m}{k} \quad \text { or } \quad m=\frac{T^2 k}{4 \pi^2} \\
& m=\frac{0.6 \times 0.6 \times 2460 \times 49}{4 \times 484} \mathrm{~kg}=22.3 \mathrm{~kg} \\
& \Rightarrow \quad m g=22.3 \times 9.8 \mathrm{~N}=218.5 \mathrm{~N}=22.3 \mathrm{kgf} \text {. } \\
&
\end{aligned}
\)

Hint

Here, \(\quad \mathrm{K}=1200 \mathrm{Nm}^{-1} ; \mathrm{m}=3.0 \mathrm{~kg}, a=2.0 \mathrm{~cm}=0.02 \mathrm{~m}\)
(i) Frequency, \(\quad v=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}}=3.2 \mathrm{~s}^{-1}\)
(ii) Acceleration, \(\quad \mathrm{A}=\omega^2 \quad y=\frac{k}{m} y\)
Acceleration will be maximum when \(y\) is maximum i.e., \(y=a\)
\(\therefore\) max. acceleration, \(\mathrm{A}_{\max }=\frac{k a}{m}=\frac{1200 \times 0.02}{3}=8 \mathrm{~ms}^{-2}\)
(iii) Max. speed of the mass will be when it is passing through mean position
\(
V_{\max }=a \omega=a \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}}=0.4 \mathrm{~ms}^{-1}
\)

Hint

\(
a=2 \mathrm{~cm}, \quad \omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{1200}{3}} \mathrm{~s}^{-1}=20 \mathrm{~s}^{-1}
\)
(a) Since time is measured from mean position,
\(
x=a \sin \omega t=2 \sin 20 t
\)
(b) At the maximum stretched position, the body is at the extreme right position. The initial phase is \(\frac{\pi}{2}\).
\(
\therefore \quad x=a \sin \left(\omega t+\frac{\pi}{2}\right)=a \cos \omega t=2 \cos 20 t
\)
(c) At the maximum compressed position, the body is at the extreme left position. The initial phase is \(\frac{3 \pi}{2}\).
\(
\therefore \quad x=a \sin \left(\omega t+\frac{3 \pi}{2}\right)=-a \cos \omega t=-2 \cos 20 t
\)
Note: The functions neither differ in amplitude nor in frequency. They differ only in the initial phase.

Hint

(a) Let \(\mathrm{A}\) be any point on the circle of reference of the fig. (a) From A, draw BN perpendicular on \(x\)-axis.
\(
\text { If } \quad \begin{aligned}
& \angle P O A=\theta, \text { then } \\
& \angle O A M=\theta=\omega t
\end{aligned} 
\)
\(\therefore\) In triangle OAM,
\(
\begin{aligned}
& \frac{O M}{O A} & =\sin \theta \\
\therefore & \frac{-x}{3} & =\sin \omega t=\sin \frac{2 \pi}{T} t \\
\therefore & x & =-3 \sin \frac{2 \pi}{2} t \text { or } x=-3 \sin \pi t \\
\end{aligned}
\)
which is the equation of SHM.

(b) (b) Let \(B\) be any point on the circle of reference of fig. (b). From \(B\), draw \(B N\) perpendicular on \(x\)-axis.
Then \(\quad \angle B O N=\theta=\omega t\)
\(
\begin{array}{rlrl}
\therefore \text { In } \triangle O N B, \quad \cos \theta & =\frac{O N}{O B} \\
& \text { or } \quad O N & =O B \cos \theta
\end{array}
\)
\(
\begin{aligned}
\therefore & -x & =2 \cos \omega t \\
\Rightarrow & x & =-2 \cos \frac{2 \pi}{T} t=-2 \cos \frac{2 \pi}{4} t
\end{aligned}
\)
\(\therefore \quad x=-2 \cos \frac{\pi}{4} t \quad\) which is equation of SHM.

Hint

(a) Let \(y\) be the maximum extension produced in the spring in Fig. (a)
Then \(\quad F=k y\) (in magnitude) \(\therefore y=\frac{F}{k}\)
If fig. (b), the force on one mass acts as the force of reaction due to the force on the other mass. Therefore, each mass behaves as if it is fixed with respect to the other.
Therefore, \(\quad F=k y \Rightarrow y=\frac{F}{k}\)
(b) In fig. \((a), \quad F=-k y\)
\(\Rightarrow \quad m a=-k y \Rightarrow a=-\frac{k}{m} y \quad \therefore \quad \omega^2=\frac{k}{m} \quad\) i.e., \(\quad \omega=\sqrt{\frac{k}{m}}\)
Therefore, period \(T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{k}}\)
In fig. (b), we may consider that the centre of the system is \(\mathrm{O}\) and there are two springs each of length \(\frac{l}{2}\) attached to the two masses, each \(m\), so that \(k^{\prime}\) is the spring factor of each of the springs.
Then,
\(
K^{\prime}=2 k
\)
\(
\begin{aligned}
\therefore \quad T & =2 \pi \sqrt{\frac{m}{k^{\prime}}} \\
& =2 \pi \sqrt{\frac{m}{2 k}}
\end{aligned}
\)

Hint

Stroke of piston \(=2\) times the amplitude
Let \(\mathrm{A}=\) amplitude, stroke \(=1 \mathrm{~m}\)
\(
\therefore \quad \Rightarrow A=\frac{1}{2} \mathrm{~m}
\)
Angular frequency, \(\omega=200 \mathrm{rad} / \mathrm{min}\).
\(
V_{\max }=?
\)
We know that the maximum speed of the block when the amplitude is \(A\),
\(
V_{\max }=\omega \mathrm{A}=200 \times \frac{1}{2}=100 \mathrm{~m} / \mathrm{min} .
\)

Hint

Here, \(\quad g_m=1.7 \mathrm{~ms}^{-2} ; g_e=9.8 \mathrm{~ms}^{-2} ; T_m=? ; \quad T_e=3.5 \mathrm{~s}^{-1}\)
Since,
\(
T_e=2 \pi \sqrt{\frac{1}{g_e}} \text { and } T_m=2 \pi \sqrt{\frac{1}{g_m}}
\)
\(
\begin{aligned}
\therefore \quad \frac{T_m}{T_e} & =\sqrt{\frac{g_e}{g_m}} \Rightarrow T_m=T_e=\sqrt{\frac{g_e}{g_m}} \\
& =3.5 \sqrt{\frac{9.8}{1.7}}=8.4 \mathrm{~s} .
\end{aligned}
\)

Hint

In this case, the bob of the pendulum is under the action of two accelerations.
(i) Acceleration due to gravity ‘ \(g\) ‘ acting vertically downwards.
(ii) Centripetal acceleration \(a_c=\frac{v^2}{R}\) acting along the horizontal direction.
\(\therefore\) Effective acceleration, \(g^{\prime}=\sqrt{g^2+a_c^2}\)
or
\(
g^{\prime}=\sqrt{g^2+\frac{v^4}{R^2}}
\)
Now time period, \(T^{\prime}=2 \pi \sqrt{\frac{1}{g}}=2 \pi \sqrt{\frac{l}{\sqrt{g^2+\frac{v^4}{R^2}}}}\).

Hint

(a) Here, mass, \(\mathrm{M}=300 \mathrm{~kg}\), displacement, \(x=15 \mathrm{~cm}=0.15 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2\). There are four spring systems. If \(k\) is the spring constant of each spring, then total spring constant of all the four springs in parallel is
\(
\begin{aligned}
K_p & =4 k \quad \therefore \quad M_g=k_p x=4 k x \\
\Rightarrow \quad K & =\frac{M g}{4 x}=\frac{3000 \times 10}{4 \times 0.15}=5 \times 10^4 \mathrm{~N} .
\end{aligned}
\)
(b) For each spring system supporting \(750 \mathrm{~kg}\) of weight,
\(\therefore\) Using \(\quad x=x_0 e^{-\frac{b t}{2 m}}\), we get
\(
\frac{50}{100} x_0=x_0 e^{-\frac{b \times 0.77}{2 \times 750}} \text { or } e^{\frac{0.77 b}{1500}}=2
\)
Taking logarithm of both sides,
\(
\begin{aligned}
\frac{0.77 b}{1500} & =\ln 2=2.303 \log 2 \\
b & =\frac{1500}{0.77} \times 2.303 \times 0.3010 \\
& =1350.4 \mathrm{~kg} \mathrm{~s}^{-1}
\end{aligned}
\)

Hint

Let the particle executing SHM starts oscillating from its mean position. Then displacement equation is
\(\begin{aligned} x & =A \sin \omega t \\ \therefore \quad \text { Particle velocity, } v & =A \omega \cos \omega t \\ \therefore \quad \text { Instantaneous K.E., } K & =\frac{1}{2} m v^2=\frac{1}{2} m A^2 \omega^2 \cos ^2 \omega t\end{aligned}\)
\(\therefore\) Average value of K.E. over one complete cycle
\(
K_{a v}=\frac{1}{T} \int_0^T \frac{1}{2} m A^2 \omega^2 \cos ^2 \omega t d t=\frac{m A^2 \omega^2}{2 T} \int_0^T \cos ^2 \omega t d t
\)
\(
\begin{aligned}
& =\frac{m A^2 \omega^2}{2 T} \int_0^T \frac{(1+\cos 2 \omega t)}{2} d t \\
& =\frac{m A^2 \omega^2}{4 T}\left[t+\frac{\sin 2 \omega t}{2 \omega}\right]_0^T \\
& =\frac{m A^2 \omega^2}{4 T}\left[(T-0)+\left(\frac{\sin 2 \omega t-\sin 0}{2 \omega}\right)\right] \\
& =\frac{1}{4} m A^2 w^2 \dots(i)
\end{aligned}
\)
Again instantaneous P.E., \(\mathrm{U}=\frac{1}{2} k x^2=\frac{1}{2} m \omega^2 x^2=\frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t\)
\(\therefore\) Average value of P.E. over one complete cycle
\(
\begin{aligned}
U_{a v} & =\frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t=\frac{m \omega^2 A^2}{2 T} \int_0^T \sin ^2 \omega t d t \\
& =\frac{m \omega^2 A^2}{2 T} \int_0^T \frac{(1-\cos 2 \omega t)}{2} d t \\
& =\frac{m \omega^2 A^2}{4 T}\left[t-\frac{\sin 2 \omega t}{2 \omega}\right]_0^T \\
& =\frac{m \omega^2 A^2}{4 T}\left[(T-0)-\frac{(\sin 2 \omega t-\sin 0)}{2 \omega}\right] \\
& =\frac{1}{4} m \omega^2 A^2 \dots(ii)
\end{aligned}
\)
Simple comparison of (i) and (ii), shows that
\(
K_{a v}=U_{a v}=\frac{1}{4} m \omega^2 A^2
\)

Hint

\(
\begin{aligned}
T & =2 \pi \sqrt{\frac{I}{\alpha}} \quad \text { or } \quad T^2=\frac{4 \pi^2 \mathrm{I}}{\alpha} \\
\alpha & =\frac{4 \pi^2 \mathrm{I}}{T^2} \text { or } \alpha=\frac{4 \pi^2}{T^2}\left(\frac{1}{2} M R^2\right) \\
\alpha & =\frac{2 \pi^2 M R^2}{T^2} \\
\alpha & =\frac{2(3.14)^2 \times 10 \times(0.15)^2}{(1.5)^2} \mathrm{Nm} \mathrm{rad}^{-1} \\
& =1.97 \mathrm{Nm} \mathrm{rad}^{-1} \approx 2.0\mathrm{Nm} \mathrm{rad}^{-1} .
\end{aligned}
\)

Hint

Here, \(r=5 \mathrm{~cm}=0.05 \mathrm{~m} ; \mathrm{T}=0.2 \mathrm{~s} ; \omega=\frac{2 \pi}{T}=\frac{2 \pi}{0.2}=10 \pi \mathrm{rad} / \mathrm{s}\)
When displacement is \(y\), then
acceleration, \(A=-\omega^2 y\)
velocity, \(V=\omega \sqrt{r^2-y^2}\)

Here, \(r=5 \mathrm{~cm}=0.05 \mathrm{~m} ; \mathrm{T}=0.2 \mathrm{~s} ; \omega=\frac{2 \pi}{T}=\frac{2 \pi}{0.2}=10 \pi \mathrm{rad} / \mathrm{s}\)
When displacement is \(y\), then
acceleration, \(A=-\omega^2 y\)
velocity, \(V=\omega \sqrt{r^2-y^2}\)
Case (a) When
\(
\begin{aligned}
y & =5 \mathrm{~cm}=0.05 \mathrm{~m} \\
A & =-(10 \pi)^2 \times 0.05=-5 \pi^2 \mathrm{~m} / \mathrm{s}^2 \\
V & =10 \pi \sqrt{(0.05)^2-(0.05)^2}=0
\end{aligned}
\)
Case (b) When
\(
\begin{aligned}
& y=3 \mathrm{~cm}=0.03 \mathrm{~m} \\
& A=-(10 \pi)^2 \times 0.03=-3 \pi^2 \mathrm{~m} / \mathrm{s}^2 \\
& V=10 \pi \sqrt{(0.05)^2-(0.03)^2}=10 \pi \times 0.04=0.4 \pi \mathrm{m} / \mathrm{s}
\end{aligned}
\)
Case (c) When
\(
\begin{aligned}
& y=0, \quad \mathrm{~A}=-(10 \pi)^2 \times 0=0 \\
& V=10 \pi \sqrt{(0.05)^2-0^2}=10 \pi \times 0.05=0.5 \pi \mathrm{m} / \mathrm{s}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x=\alpha \cos (\omega t+\theta) \\
& v=\frac{d x}{d t}=-a \omega \sin (\omega t+\theta)
\end{aligned}
\)
\(
\begin{array}{ll}
\text { When } & t=0, x=x_0 \text { and } \frac{d x}{d t}=-v_0 \\
\therefore & x_0=a \cos \theta \dots(i)
\end{array}
\)
and
\(
-v_0=-a \omega \sin \theta \text { or } a \sin \theta=\frac{v_0}{\omega} \dots(ii)
\)
Squaring and adding ( \(i\) ) and (ii), we get
\(
\begin{aligned}
a^2\left(\cos ^2 \theta+\sin ^2 \theta\right) & =x_0^2+\frac{v_0^2}{\omega^2} \\
a & =\sqrt{x_0^2+\frac{v_0^2}{\omega^2}}
\end{aligned}
\)

Hint

Comparing with equation \(x=A \sin (\omega t+\delta)\), we see that
the amplitude \(=5 \mathrm{~m}\),
and time period \(=\frac{2 \pi}{\omega}=\frac{2 \pi}{\pi \mathrm{s}^{-1}}=2 \mathrm{~s}\).
The maximum speed \(=A \omega=5 \mathrm{~m} \times \pi \mathrm{s}^{-1}=5 \pi \mathrm{ms}^{-1}\).
The velocity at time \(t=\frac{d x}{d t}=A \omega \cos (\omega t+\delta)\).
At \(t=1 \mathrm{~s}\)
\(
v=(5 \mathrm{~m})\left(\pi \mathrm{s}^{-1}\right) \cos \left(\pi+\frac{\pi}{3}\right)=-\frac{5 \pi}{2} \mathrm{~ms}^{-1}
\)

Hint

The maximum force exerted on the block is \(k A\) when the block is at the extreme position.
The angular frequency \(\omega=\frac{2 \pi}{T}=2 \mathrm{~s}^{-1}\).
The spring constant \(\quad=k=m \omega^2\)
\(=(5 \mathrm{~kg})\left(4 \mathrm{~s}^{-2}\right)=20 ~\mathrm{Nm}^{-1}\).
Maximum force \(=k A=\left(20 ~\mathrm{Nm}^{-1}\right)\left(0^{\cdot} 1 \mathrm{~m}\right)=2 \mathrm{~N}\).

Hint

(a)
\(
\text { Time period }=\frac{2 \pi}{\omega}=\frac{2 \pi}{6 \cdot 28} \mathrm{~s}=1 \mathrm{~s} \text {. }
\)
(b)
\(
\begin{aligned}
\text { Maximum speed } & =A \omega=(0.1 \mathrm{~m})\left(6.28 \mathrm{~s}^{-1}\right) \\
& =0.628 \mathrm{~m} \mathrm{~s}^{-1} .
\end{aligned}
\)
(c)
\(
\begin{aligned}
\text { Maximum acceleration } & =A \omega^2 \\
& =(0 \cdot 1 \mathrm{~m})\left(6 \cdot 28 \mathrm{~s}^{-1}\right)^2 \\
& =4 \mathrm{~m} \mathrm{~s}^{-2} .
\end{aligned}
\)
(d)
\(
\begin{aligned}
v=\omega \sqrt{A^2-x^2} & =\left(6 \cdot 28 \mathrm{~s}^{-1}\right) \sqrt{(10 \mathrm{~cm})^2-(6 \mathrm{~cm})^2} \\
& =50^{-} 2 \mathrm{~cm} \mathrm{~s}^{-1} .
\end{aligned}
\)
(e)
\(
\text { At } t=0 \text {, the velocity is zero, i.e., the particle is at an extreme.}
\)
The equation for displacement may be written as
\(
x=A \cos \omega t
\)
\(
\text { The velocity is } v=-A \omega \sin \omega t \text {. }
\)
\(
\text { At } t=\frac{1}{6} \mathrm{~s}, \quad v=-(0 \cdot 1 \mathrm{~m})\left(6 \cdot 28 \mathrm{~s}^{-1}\right) \sin \left(\frac{6 \cdot 28}{6}\right)
\)
\(
\begin{aligned}
& =\left(-0.628 \mathrm{~m} \mathrm{~s}^{-1}\right) \sin \frac{\pi}{3} \\
& =-54.4 \mathrm{~cm} \mathrm{~s}^{-1} .
\end{aligned}
\)

Hint

Let the equation of motion be \(x=A \sin \omega t\).
At \(t=0, x=0\) and hence the particle is at its mean position. Its velocity is
\(
v=A \omega \cos \omega t=A \omega
\)
which is positive. So it is going towards \(x=A / 2\). The particle will be at \(x=A / 2\), at a time \(t\), where \(\frac{A}{2}=A \sin \omega t\)
or, \(\quad \sin \omega t=1 / 2\)
or, \(\quad \omega t=\pi / 6\).
Here the minimum positive value of \(\omega t\) is chosen because we are interested in finding the time taken by the particle to directly go from \(x=0\) to \(x=A / 2\).
Thus, \(t=\frac{\pi}{6 \omega}=\frac{\pi}{6(2 \pi / T)}=\frac{T}{12}\).

Hint

Suppose the length of the spring is stretched by a length \(\Delta l\). The tension in the spring is \(k \Delta l\) and this is the force by the spring on the block. The other force on the block is \(m g\) due to gravity. For equilibrium, \(m g=k \Delta l\) or \(\Delta l=m g / k\). Take this position of the block as \(x=0\). If the block is further displaced by \(x\), the resultant force is \(k\left(\frac{m g}{k}+x\right)-m g=k x\).

Thus, the resultant force is proportional to the displacement. The motion is simple harmonic with a time period \(T=2 \pi \sqrt{\frac{m}{k}}\).

We see that in vertical oscillations, gravity has no effect on time period. The only effect it has is to shift the equilibrium position by a distance \(\mathrm{mg} / \mathrm{k}\) as if the natural length is increased (or decreased if the lower end of the spring is fixed) by \(m g / k\).

Hint

(a) The mean position of the particle during vertical oscillations is \(\mathrm{mg} / \mathrm{k}\) distance away from its position when the spring is unstretched. At the highest point, i.e., at an extreme position, the spring is unstretched.
Hence the amplitude is
\(
A=\frac{m g}{k} \dots(i)
\)
The angular frequency is
\(
\omega=\sqrt{\frac{k}{m}}=2 \pi v=(20 \pi) \mathrm{s}^{-1} \dots(ii)
\)
\(
\text { or, } \quad \frac{m}{k}=\frac{1}{400 \pi^2} \mathrm{~s}^2 \text {. }
\)
Putting in (i), the amplitude is
\(
\begin{aligned}
A & =\left(\frac{1}{400 \pi^2} \mathrm{~s}^2\right)\left(\pi^2 \frac{\mathrm{m}}{\mathrm{s}^2}\right) \\
& =\frac{1}{400} \mathrm{~m}=0.25 \mathrm{~cm} .
\end{aligned}
\)
The maximum speed \(=A \omega\)
\(
=(0.25 \mathrm{~cm})\left(20 \pi \mathrm{s}^{-1}\right)=5 \pi \mathrm{~cm} \mathrm{s}^{-1} .
\)
(b) When the spring is stretched by \(0.20 \mathrm{~cm}\), the block is \(0.25 \mathrm{~cm}-0.20 \mathrm{~cm}=0.05 \mathrm{~cm}\) above the mean position. The speed at this position will be
\(
\begin{aligned}
v & =\omega \sqrt{A^2-x^2} \\
& =\left(20 \pi \mathrm{s}^{-1}\right) \sqrt{(0 \cdot 25 \mathrm{~cm})^2-(0 \cdot 05 \mathrm{~cm})^2} \\
& \equiv 15.4 \mathrm{~cm} \mathrm{~s}^{-1} .
\end{aligned}
\)

Hint

Let us first find the equilibrium position. For the rotational equilibrium of the pulley, the tensions in the two strings should be equal. Only then the torque on the pulley will be zero. Let this tension be \(T\). The extension of the spring will be \(y=T / k\), as the tension in the spring will be the same as the tension in the string. For translational equilibrium of the pulley,
\(
2 T=m g \text { or }, 2 k y=m g \quad \text { or, } \quad y=\frac{m g}{2 k} \text {. }
\)
The spring is extended by a distance \(\frac{m g}{2 k}\) when the pulley is in equilibrium.
Now suppose, the centre of the pulley goes down further by a distance \(x\). The total increase in the length of the string plus the spring is \(2 x\) ( \(x\) on the left of the pulley and \(x\) on the right). As the string has a constant length, the extension of the spring is \(2 x\). The energy of the system is
\(
U=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2-m g x+\frac{1}{2} k\left(\frac{m g}{2 k}+2 x\right)^2
\)
\(
=\frac{1}{2}\left(\frac{I}{r^2}+m\right) v^2+\frac{m^2 g^2}{8 k}+2 k x^2 \text {. }
\)
As the system is conservative, \(\frac{d U}{d t}=0\), giving \(0=\left(\frac{I}{r^2}+m\right) v \frac{d v}{d t}+4 k x v\) or, \(\quad \frac{d v}{d t}=-\frac{4 k x}{\left(\frac{I}{r^2}+m\right)}\)
or, \(\quad a=-\omega^2 x\), where \(\omega^2=\frac{4 k}{\left(\frac{I}{r^2}+m\right)}\).
Thus, the centre of mass of the pulley executes a simple harmonic motion with time period
\(
T=2 \pi \sqrt{\left(\frac{I}{r^2}+m\right) /(4 k)}
\)

Hint

(a) For small amplitude, the two blocks oscillate together. The angular frequency is
\(
\omega=\sqrt{\frac{k}{M+m}}
\)
and so the time period \(T=2 \pi \sqrt{\frac{M+m}{k}}\).
(b) The acceleration of the blocks at displacement \(x\) from the mean position is
\(
a=-\omega^2 x=\frac{-k x}{M+m} .
\)
The resultant force on the upper block is, therefore,
\(
m a=\frac{-m k x}{M+m} \text {. }
\)
This force is provided by the friction of the lower block. Hence, the magnitude of the frictional force is \(\frac{m k|x|}{M+m}\).
(c) Maximum force of friction required for simple harmonic motion of the upper block is \(\frac{m k A}{M+m}\) at the
extreme positions. But the maximum frictional force can only be \(\mu \mathrm{mg}\). Hence
\(
\frac{m k A}{M+m}=\mu m g \text { or, } A=\frac{\mu(M+m) g}{k}
\)

Hint

Assuming the collision to last for a small interval only, we can apply the principle of conservation of momentum. The common velocity after the collision is \(\frac{v}{2}\). The kinetic energy \(=\frac{1}{2}(2 m)\left(\frac{v}{2}\right)^2=\frac{1}{4} m v^2\). This is also the total energy of vibration as the spring is unstretched at this moment. If the amplitude is \(A\), the total energy can also be written as \(\frac{1}{2} k A^2\). Thus
\(
\frac{1}{2} k A^2=\frac{1}{4} m v^2, \text { giving } A=\sqrt{\frac{m}{2 k}} v 
\)

Hint

Considering “the two blocks plus the spring” as a system, there is no external resultant force on the system. Hence the centre of mass of the system will remain at rest. The mean positions of the two simple harmonic motions occur when the spring becomes unstretched. If the mass \(m\) moves towards right through a distance \(x\) and the mass \(M\) moves towards left through a distance \(X\) before the spring acquires natural length,
\(
x+X=x_0 \dots(i)
\)
\(x\) and \(X\) will be the amplitudes of the two blocks \(m\) and \(M\) respectively. As the centre of mass should not change during the motion, we should also have
\(
m x=M X \dots(ii)
\)
From (i) and (ii), \(x=\frac{M x_0}{M+m}\) and \(X=\frac{m x_0}{M+m}\).
Hence, the left block is \(x=\frac{M x_0}{M+m}\) distance away from its mean position in the beginning of the motion. The force by the spring on this block at this instant is equal to the tension of spring, i.e., \(T=k x_0\).
Now \(x=\frac{M x_0}{M+m}\) or, \(x_0=\frac{M+m}{M} x\)
Thus, \(T=\frac{k(M+m)}{M} x\) or, \(a=\frac{T}{m}=\frac{k(M+m)}{M m} x\).
The angular frequency is, therefore, \(\omega=\sqrt{\frac{k(M+m)}{M m}}\) and the frequency is \(\mathrm{v}=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{k(M+m)}{M m}}\).

Hint

Consider the situation shown in figure below. Suppose at an instant \(t\) the particle in the tunnel is at a distance \(x\) from the centre of the earth. Let us draw a sphere of radius \(x\) with its centre at the centre of the earth. Only the part of the earth within this sphere will exert a net attraction on the particle. Mass of this part is
\(
M^{\prime}=\frac{\frac{4}{3} \pi x^3}{\frac{4}{3} \pi R^3} M=\frac{x^3}{R^3} M
\)
The force of attraction is, therefore,
\(
F=\frac{G\left(x^3 / R^2\right) M m}{x^2}=\frac{G M m}{R^3} x .
\)
This force acts towards the centre of the earth. Thus, the resultant force on the particle is opposite to the displacement from the centre of the earth and is proportional to it. The particle, therefore, executes a simple harmonic motion in the tunnel with the centre of the earth as the mean position.

The force constant is \(k=\frac{G M m}{R^3}\), so that the time period is
\(
T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{R^3}{G M}} .
\)

Hint

(a) The angular frequency is
\(
\omega=\sqrt{g / l}=\sqrt{\frac{10 \mathrm{~m} \mathrm{~s}^{-2}}{0 \cdot 4 \mathrm{~m}}}=5 \mathrm{~s}^{-1}
\)
The time period is
\(
\frac{2 \pi}{\omega}=\frac{2 \pi}{5 \mathrm{~s}^{-1}}=1 \cdot 26 \mathrm{~s}
\)
\(
\text { (b) Linear amplitude }=40 \mathrm{~cm} \times 0.04=1.6 \mathrm{~cm}
\)
(c) Angular speed at displacement \(0.02 \mathrm{~rad}\) is
\(
\Omega=\left(5 \mathrm{~s}^{-1}\right) \sqrt{(0.04)^2-(0.02)^2} \mathrm{~rad}=0.17 \mathrm{~rads}^{-1}
\)
Linear speed of the bob at this instant
\(
=(40 \mathrm{~cm}) \times 0.17 \mathrm{~s}^{-1}=6.8 \mathrm{~cm} \mathrm{~s}^{-1} .
\)
(d) At momentary rest, the bob is in extreme position. Thus, the angular acceleration
\(
\alpha=(0.04 \mathrm{~rad})\left(25 \mathrm{~s}^{-2}\right)=1 \mathrm{~rad} \mathrm{s}^{-2}
\)

Hint

Suppose the speed of the bob at angle \(\theta\) is \(v\). Using conservation of energy between the extreme position and the position with angle \(\theta\),
\(
\frac{1}{2} m v^2=m g l\left(\cos \theta-\cos \theta_0\right) \dots(i)
\)

As the bob moves in a circular path, the force towards the centre should be equal to \(m v^2 / l\). Thus,
\(
T-m g \cos \theta=m v^2 / l
\)
Using (i),
\(
\begin{aligned}
& T-m g \cos \theta=2 m g\left(\cos \theta-\cos \theta_0\right) \\
& \text { or, } \quad T=3 m g \cos \theta-2 m g \cos \theta_0 .
\end{aligned}
\)
Now \(\cos \theta\) is maximum at \(\theta=0\) and decreases as \(|\theta|\) increases (for \(|\theta|<90^{\circ}\) ).
Thus, the tension is maximum when \(\theta=0\), i.e., at the mean position and is minimum when \(\theta= \pm \theta_0\), i.e., at extreme positions.

Hint

At a height \(R\) (radius of the earth) the acceleration due to gravity is
\(
g^{\prime}=\frac{G M}{(R+R)^2}=\frac{1}{4} \frac{G M}{R^2}=g / 4
\)
The time period of small oscillations of the simple pendulum is
\(
T=2 \pi \sqrt{l / g^{\prime}}=2 \pi \sqrt{\frac{1 \cdot 0 \mathrm{~m}}{\frac{1}{4} \times \pi^2 \mathrm{~ms}^{-2}}}=2 \pi\left(\frac{2}{\pi} \mathrm{s}\right)=4 \mathrm{~s} .
\)

Hint

Let the mass of the stick be \(m\). The moment of inertia of the stick about the axis of rotation through the point of suspension is
\(
I=\frac{m l^2}{12}+m d^2,
\)
where \(l=1 \mathrm{~m}\) and \(d=40 \mathrm{~cm}\).
The separation between the centre of mass of the stick and the point of suspension is \(d=40 \mathrm{~cm}\).

The time period of this physical pendulum is
\(
\begin{aligned}
T & =2 \pi \sqrt{\frac{I}{m g d}} \\
& =2 \pi \sqrt{\left(\frac{m l^2}{12}+m d^2\right) /(m g d)} \\
& =2 \pi\left[\sqrt{\left(\frac{1}{12}+0.16\right) / 4}\right] \mathrm{s}=1.55 \mathrm{s.}
\end{aligned}
\)

Hint

Let the torsional constant of the wire be \(k\). The moment of inertia of the first disc about the wire is \(0.2 \mathrm{~kg}-\mathrm{m}^2\). Hence, the time period is
\(
\begin{aligned}
2 \mathrm{~s} & =2 \pi \sqrt{\frac{I}{K}} \\
& =2 \pi \sqrt{\frac{0 \cdot 2 \mathrm{~kg}-\mathrm{m}^2}{k}} \dots(i)
\end{aligned}
\)
When the second disc having moment of inertia \(I_1\) about the wire is added, the time period is
\(
2.5 \mathrm{~s}=2 \pi \sqrt{\frac{0.2 \mathrm{~kg}-\mathrm{m}^2+I_1}{k}} \dots(ii)
\)
From (i) and (ii), \(\frac{6 \cdot 25}{4}=\frac{0 \cdot 2 \mathrm{~kg}-\mathrm{m}^2+I_1}{0 \cdot 2 \mathrm{~kg}-\mathrm{m}^2}\).
This gives \(I_1 \approx 0.11 \mathrm{~kg}-\mathrm{m}^2\).

Hint

(a) The oscillations take place about the horizontal line through the point of suspension and perpendicular to the plane of the figure. The moment of inertia of the rod about this line is
\(
\begin{aligned}
& \frac{m l^2}{12}+m l^2=\frac{13}{12} m l^2 . \\
& \text { The time period }=2 \pi \sqrt{\frac{I}{m g l}}=2 \pi \sqrt{\frac{13 m l^2}{12 m g l}} \\
& =2 \pi \sqrt{\frac{13 l}{12 g}} . \\
&
\end{aligned}
\)
(b) The angular oscillations take place about the suspension wire. The moment of inertia about this line is \(m l^2 / 12\). The time period is
\(
2 \pi \sqrt{\frac{I}{k}}=2 \pi \sqrt{\frac{m l^2}{12 k}} .
\)

Hint

\(
\text { (a) At } t=0, \quad x_1=A_1 \sin \omega t=0
\)
\(
\text { and } \begin{aligned}
x_2 & =A_2 \sin (\omega t+\pi / 3) \\
& =A_2 \sin (\pi / 3)=\frac{A_2 \sqrt{ } 3}{2} .
\end{aligned}
\)
Thus, the resultant displacement at \(t=0\) is
\(
x=x_1+x_2=\frac{A_2 \sqrt{ } 3}{2} .
\)
(b) The resultant of the two motions is a simple harmonic motion of the same angular frequency \(\omega\). The amplitude of the resultant motion is
\(
\begin{aligned}
A & =\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos (\pi / 3)} \\
& =\sqrt{A_1^2+A_2^2+A_1 A_2} .
\end{aligned}
\)
The maximum speed is
\(
v_{\max }=A \omega=\omega \sqrt{A_1^2+A_2^2+A_1 A_2} .
\)
(c) The maximum acceleration is
\(
a_{\max }=A \omega^2=\omega^2 \sqrt{A_1^2+A_2^2+A_1 A_2} .
\)

Hint

Let the amplitudes of the individual motions be \(A\) each. The resultant amplitude is also \(A\). If the phase difference between the two motions is \(\delta\),
\(
A=\sqrt{A^2+A^2+2 A \cdot A \cdot \cos \delta}
\)
\(
\text { or, } \quad=A \sqrt{2(1+\cos \delta)}=2 A \cos \frac{\delta}{2}
\)
\(
\text { or, } \quad \cos \frac{\delta}{2}=\frac{1}{2}
\)
or,
\(
\delta=2 \pi / 3
\)

Hint

It is given,
Amplitude of the simple harmonic motion, \(A=10 \mathrm{~cm}\)
At \(t=0\) and \(x=5 \mathrm{~cm}\)
Time period of the simple harmonic motion, \(T=6 \mathrm{~s}\)
Angular frequency \((\omega)\) is given by,
\(
\omega=\frac{2 \pi}{T}=\frac{2 \pi}{6}=\frac{\pi}{3} \sec ^{-1}
\)
Consider the equation of motion of S.H.M,
\(
Y=A \sin (\omega t+\phi) \ldots(1)
\)
where \(Y\) is displacement of the particle, and \(\phi\) is phase of the particle.
On substituting the values of \(A, t\) and \(\omega\) in equation (1), we get:
\(
\begin{aligned}
& 5=10 \sin (\omega \times 0+\phi) \\
& \Rightarrow 5=10 \sin \phi
\end{aligned}
\)
\(
\begin{aligned}
& \sin \phi=\frac{1}{2} \\
& \Rightarrow \phi=\frac{\pi}{6}
\end{aligned}
\)
\(\therefore\) Equation of displacement can be written as,
\(
x=(10 \mathrm{~cm}) \sin \left(\frac{\pi}{3} t+\frac{\pi}{6}\right)
\)
(ii) At \(t=4 \mathrm{~s}\),
\(
\begin{aligned}
& x=10 \sin \left[\frac{\pi}{3} 4+\frac{\pi}{6}\right] \\
& =10 \sin \left[\frac{8 \pi+\pi}{6}\right] \\
& =10 \sin \left(\frac{9 \pi}{6}\right) \\
& =10 \sin \left(\frac{3 \pi}{2}\right) \\
& =10 \sin \left(\pi+\frac{\pi}{2}\right) \\
& =-10 \sin \frac{\pi}{2}=-10
\end{aligned}
\)
Acceleration is given by,
\(
\begin{aligned}
& a=-\omega^2 x \\
& =\left(\frac{-\pi^2}{9}\right) \times(-10) \\
& =10.9 \approx 11 \mathrm{~cm} / \mathrm{sec}^{-2}
\end{aligned}
\)

Hint

It is given that:
Position of the particle, \(x=2 \mathrm{~cm}=0.02 \mathrm{~m}\)
Velocity of the particle, \(v=1 \mathrm{~ms}^{-1}\)
Acceleration of the particle, \(a=10 \mathrm{~ms}^{-2}\).
Let
\(\omega\) be the angular frequency of the particle.
The acceleration of the particle is given by,
\(
\begin{aligned}
& a=\omega^2 x \\
& \Rightarrow \omega=\sqrt{\frac{a}{x}}=\sqrt{\frac{10}{0.02}} \\
& =\sqrt{500}=10 \sqrt{5} \mathrm{~Hz}
\end{aligned}
\)
Time period of the motion is given as,
\(
\begin{aligned}
& T=\frac{2 \pi}{\omega}=\frac{2 \pi}{10 \sqrt{5}} \\
& =\frac{2 \times 3.14}{10 \times 2.236} \\
& =0.28 \mathrm{~s}
\end{aligned}
\)
Now, the amplitude \(A\) is calculated as,
\(
\begin{aligned}
& v=\omega \sqrt{A^2-x^2} \\
& \Rightarrow v^2=\omega^2\left(A^2-x^2\right) \\
& 1=500\left(A^2-0.0004\right) \\
& \Rightarrow A=0.0489=0.049 m \\
& \Rightarrow A=4.9 \mathrm{~cm}
\end{aligned}
\)

Hint

It is given that:
Amplitude of the particle executing simple harmonic motion, \(A=10 \mathrm{~cm}\)
To determine the distance from the mean position, where the kinetic energy of the particle is equal to its potential energy:
Let \(y\) be displacement of the particle,
\(\omega\) be the angular speed of the particle, and
A be the amplitude of the simple harmonic motion.
Equating the mathematical expressions for K.E. and P.E. of the particle, we get :
\(
\begin{aligned}
& \left(\frac{1}{2}\right) m \omega^2\left(A^2-y^2\right)=\left(\frac{1}{2}\right) m \omega^2 y^2 \\
& A^2-y^2=y^2 \\
& 2 y^2=A^2 \\
& \Rightarrow y=\frac{A}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5 \sqrt{2}
\end{aligned}
\)
The kinetic energy and potential energy of the particle are equal at a distance of \(5 \sqrt{2} \mathrm{~cm}\) from the mean position.

Hint

It is given that: Maximum speed of the particle, \(v_{\operatorname{Max}}=10 \mathrm{~cm}^{-1}\)
Maximum acceleration of the particle,
\(
a_{\operatorname{Max}}=50 \mathrm{cms}^{-2}
\)
The maximum velocity of a particle executing simple harmonic motion is given by, \(v_{\operatorname{Max}}=A \omega\)
where omega is angular frequency, and
\(A\) is amplitude of the particle.
Substituting the value of \(v_{M a x}\) in the above expression,
we get :
\(
\begin{aligned}
& A \omega=10 \ldots(1) \\
& \Rightarrow \omega^2=\frac{100}{A^2}
\end{aligned}
\)
\(
\begin{aligned}
& a_{\mathrm{Max}}=\omega^2 A=50 \mathrm{cms}^{-1} \\
& \Rightarrow \omega^2=\frac{50}{A} \ldots(2)
\end{aligned}
\)
From the equations (1) and (2), we get:
\(
\begin{aligned}
& \frac{100}{A^2}=\frac{50}{A} \\
& \Rightarrow A=2 \mathrm{~cm} \\
& \therefore \omega=\sqrt{\frac{100}{A^2}}=5 \mathrm{sec}^{-1}
\end{aligned}
\)
To determine the positions where the speed of the particle is \(8 \mathrm{~ms}^{-1}\), we may use the following formula:
\(
v^2=\omega^2\left(A^2-y^2\right)
\)
where \(y\) is distance of particle from the mean position, and
\(v\) is velocity of the particle.
On substituting the given values in the above equation, we get:
\(
\begin{aligned}
& 64=25\left(4-y^2\right) \\
& \Rightarrow \frac{64}{25}=4-y^2 \\
& \Rightarrow 4-y^2=2.56 \\
& \Rightarrow \quad y^2=1.44 \\
& \Rightarrow \quad y=\sqrt{1.44} \\
& \Rightarrow \quad y= \pm 1.2 \mathrm{~cm} \quad \text { (from the mean position) } \\
&
\end{aligned}
\)

Hint

Given: Equation of motion of the particle executing S.H.M , \(x=(2.0 \mathrm{~cm}) \sin \left[\left(100 \mathrm{~s}^{-1}\right) t+\frac{\pi}{6}\right]\)
Mass of the particle, \(m=10 g \ldots(1)\)
General equation of the particle is given by, \(x=A \sin (\omega t+\phi) \ldots(2)\)
On comparing the equations (1) and (2) we get:
(a) Amplitude, \(A\) is \(2 \mathrm{~cm}\).
Angular frequency, \(\omega\) is \(100 \mathrm{~s}^{-1}\).
Time period is calculated as ,
\(
\begin{aligned}
& T=\frac{2 \pi}{\omega}=\frac{2 \pi}{100}=\frac{\pi}{50} \mathrm{~s} \\
& =0.063 \mathrm{~s}
\end{aligned}
\)

Also, we know –
\(
T=2 \pi \sqrt{\frac{m}{k}}
\)
where \(k\) is the spring constant .
\(
\begin{aligned}
& \Rightarrow T^2=4 \pi^2 \frac{m}{k} \\
& \Rightarrow k=\frac{4 \pi^2 m}{T^2}=10^5 \text { dyne } / \mathrm{cm} \\
& =100 \mathrm{~N} / \mathrm{m}
\end{aligned}
\)
(b) At \(t=0\) and \(x=2 \mathrm{~cm}\)
\(\sin \frac{\pi}{6}=2 \times \frac{1}{2}=1 \mathrm{~cm}\) from the mean position,
We know:
\(
x=A \sin (\omega t+\phi)
\)
Using
\(v=\frac{\mathrm{dx}}{\mathrm{dt}}\), we get:
\(
\begin{aligned}
& v=A \omega \cos (\omega t+\phi) \\
& =2 \times 100 \cos \left(0+\frac{\pi}{6}\right) \\
& =200 \times \frac{\sqrt{3}}{2} \\
& =100 \sqrt{3} \mathrm{cms}^{-1} \\
& =1.73 \mathrm{~ms}^{-1}
\end{aligned}
\)
(c) Acceleration of the particle is given by,
\(
\begin{aligned}
a= & -\omega^2 x \\
& =100^2 \times 1=10000 \mathrm{~cm} / \mathrm{s}^2
\end{aligned}
\)

Hint

Given: The equation of motion of a particle executing S.H.M. is,
\(
x=5 \sin \left(20 t+\frac{\pi}{3}\right)
\)
The general equation of S..H.M. is give by,
\(
x=A \sin (\omega t+\phi)
\)
(a) Maximum displacement from the mean position is equal to the amplitude of the particle. As the velocity of the particle is zero at extreme position, it is at rest.
\(\therefore\) Displacement \(x=5\), which is also the amplitude of the particle.
\(
\Rightarrow 5=5 \sin \left(20 t+\frac{\pi}{3}\right)
\)
Now,
\(
\begin{aligned}
& \sin \left(20 t+\frac{\pi}{3}\right)=1=\sin \frac{\pi}{2} \\
& \Rightarrow 20 t+\frac{\pi}{3}=\frac{\pi}{2} \\
& \Rightarrow t=\frac{\pi}{120} s
\end{aligned}
\)
The particle will come to rest at \(\frac{\pi}{120} s\)
(b) Acceleration is given as,
\(
\begin{aligned}
& a=\omega^2 x \\
= & \omega^2\left[5 \sin \left(20 t+\frac{\pi}{3}\right)\right]
\end{aligned}
\)
For \(a=0\),
\(
\begin{aligned}
& 5 \sin \left(20 t+\frac{\pi}{3}\right)=0 \\
& \Rightarrow \sin \left(20 t+\frac{\pi}{3}\right)=\sin \pi \\
& \Rightarrow 20 t=\pi-\frac{\pi}{3}=\frac{2 \pi}{3} \\
& \Rightarrow t=\frac{\pi}{30} s
\end{aligned}
\)
(c) The maximum speed \((v)\) is given by,
\(
\begin{aligned}
& v=A \omega \cos \left(\omega t+\frac{\pi}{3}\right) \\
& \text { (using } v=\frac{\mathrm{dx}}{\mathrm{dt}} \text { ) } \\
& =20 \times 5 \cos \left(20 t+\frac{\pi}{3}\right)
\end{aligned}
\)
For maximum velocity :
\(
\begin{aligned}
& \cos \left(20 t+\frac{\pi}{3}\right)=-1=\cos \pi \\
& \Rightarrow 20 t=\pi-\frac{\pi}{3}=\frac{2 \pi}{3} \\
& \Rightarrow t=\frac{\pi}{30} s
\end{aligned}
\)

Hint

It is given that a particle executes S.H.M.
Equation of S.H.M. of the particle:
\(
\begin{aligned}
& x=2.0 \cos \left(50 \pi t+\tan ^{-1} 0.75\right) \\
& =2.0 \cos (50 \pi t+0.643)
\end{aligned}
\)
(a) Velocity of the particle is given by,
\(
\begin{aligned}
& v=\frac{d x}{d t} \\
& v=-100 \pi \sin (50 \pi t+0.643)
\end{aligned}
\)
As the particle comes to rest, its velocity becomes be zero.
\(
\begin{aligned}
& \Rightarrow v=-100 \pi \sin (50 \pi t+0.643)=0 \\
& \Rightarrow \sin (50 \pi t+0.643)=0=\sin \pi
\end{aligned}
\)
When the particle initially comes to rest,
\(
\begin{aligned}
& 50 \pi t+0.643=\pi \\
& \Rightarrow t=1.6 \times 10^{-2} \mathrm{~s}
\end{aligned}
\)
(b) Acceleration is given by,
\(
\begin{aligned}
& a=\frac{d v}{d t} \\
& =-100 \pi \times 50 \pi \cos (50 \pi t+0.643)
\end{aligned}
\)
For maximum acceleration:
\(
\begin{aligned}
& \cos (50 \pi \mathrm{t}+0.643)=-1=\cos \pi(\max ) \quad \text { (so that } a \text { is } \max \text { ) } \\
& \Rightarrow \quad t=1.6 \times 10^{-2} \mathrm{~s}
\end{aligned}
\)
(c) When the particle comes to rest for the second time, the time is given as,
\(
\begin{aligned}
& 50 \pi \mathrm{t}+0.643=2 \pi \\
& \Rightarrow t=3.6 \times 10^{-2} \mathrm{~s}
\end{aligned}
\)

Hint

As per the conditions given in the question,
\(
\begin{aligned}
y_1 & =\frac{A}{2} \\
y_2 & =A
\end{aligned}
\)
(for the given two positions)
Let \(y_1\) and \(y_2\) be the displacements at the two positions and \(A\) be the amplitude.
Equation of motion for the displacement at the first position is given by,
\(
y_1=A \sin \omega t_1
\)
As displacement is equal to the half of the amplitude,
\(
\begin{aligned}
& \frac{A}{2}=A \sin \omega t_1 \\
& \Rightarrow \sin \omega t_1=\frac{1}{2} \\
& \Rightarrow \frac{2 \pi \times t_1}{T}=\frac{\pi}{6} \\
& \Rightarrow t_1=\frac{T}{12}
\end{aligned}
\)
The displacement at second position is given by,
\(
y_2=A \sin \omega t_2
\)
As displacement is equal to the amplitude at this position,
\(
\begin{aligned}
& \Rightarrow \quad A=A \sin \omega t_2 \\
& \Rightarrow \sin \omega t_2=1 \\
& \Rightarrow \omega t_2=\frac{\pi}{2} \\
& \Rightarrow\left(\frac{2 \pi}{T}\right) t_2=\frac{\pi}{2}\left(\because \sin \frac{\pi}{2}=1\right) \\
& \Rightarrow t_2=\frac{T}{4} \\
& \therefore t_2-t_1=\frac{T}{4}-\frac{T}{12}=\frac{T}{6}
\end{aligned}
\)

Hint

Given:
Spring constant, \(k=0.1 \mathrm{~N} / \mathrm{m}\)
Time period of the pendulum of clock, \(T=2 \mathrm{~s}\)
Mass attached to the string, \(m\), is to be found.
The relation between time period and spring constant is given as,
\(
T=2 \pi \sqrt{\left(\frac{m}{k}\right)}
\)
On substituting the respective values, we get:
\(
\begin{aligned}
& 2=2 \pi \sqrt{\frac{m}{k}} \\
& \Rightarrow \pi^2\left(\frac{m}{0.1}\right)=1 \\
& \therefore m=\frac{0.1}{\pi^2}=\frac{0.1}{10} \\
& =0.01 \mathrm{~kg} \approx 10 \mathrm{~g}
\end{aligned}
\)

Hint

An equivalent simple pendulum has same time period as that of the spring mass system. The time period of a simple pendulum is given by,
\(
T_p=2 \pi \sqrt{\left(\frac{l}{g}\right)}
\)
where \(l\) is the length of the pendulum, and
\(g\) is acceleration due to gravity.
Time period of the spring is given by,
\(
T_s=2 \pi \sqrt{\left(\frac{m}{k}\right)}
\)
where \(m\) is the mass, and
\(k\) is the spring constant.
Let \(x\) be the extension of the spring.
For small frequency, \(T_{\mathrm{P}}\) can be taken as equal to \(T_{\mathrm{S}}\).
\(
\begin{aligned}
& \Rightarrow \sqrt{\left(\frac{l}{g}\right)}=\sqrt{\left(\frac{m}{k}\right)} \\
& \Rightarrow\left(\frac{l}{g}\right)=\left(\frac{m}{k}\right) \\
& \Rightarrow l=\frac{m g}{k}=\frac{F}{k}=x \\
& (\because \text { restoring force }=\text { weight }=m g) \\
& \therefore l=x(\text { proved })
\end{aligned}
\)