Numerical Problems-2 - Chapter 10 Oscillations 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum oscillator. The acceleration of the bob of the pendulum is $20 \mathrm{~ms}^{-2}$ at a distance of $5 \mathrm{~m}$ from the mean position. To find the time period of oscillation.
Answer:
Given; $\mathrm{a}=20 \mathrm{~ms}^{-2} ; \mathrm{y}=5 \mathrm{ma}=\omega^2 \mathrm{y}$
$
20=\omega^2(5) ; \omega^2=4
$

Time period of oscillation,
$
\begin{aligned}
20 & =\omega^2(5) ; \omega^2 \\
\omega & =2 \mathrm{rad} / \mathrm{sec} \\
\mathrm{T} & =\frac{2 \pi}{\omega}=\frac{2 \pi}{2} \\
\mathrm{~T} & =\pi \mathrm{sec}
\end{aligned}
$

Question 2.
The acceleration due to gravity on the surface of the moon is $1.7 \mathrm{~ms}^{-2}$. What is the time period of simple pendulum on the moon if its time period on the earth is $3.5 \mathrm{~s}$ ? Give $g$ on Earth $=9.8 \mathrm{~ms}^{-2}$

Answer:
For the moon:
For the earth :
But,
$
\begin{aligned}
g_{\mathrm{m}} & =1.7 \mathrm{~ms}^{-2}, \mathrm{~T}_m=? \\
g_e & =9.8 \mathrm{~ms}^{-2}, \mathrm{~T}_e=3.5 \mathrm{~s} \\
\mathrm{~T}_e & =2 \pi \sqrt{\frac{l}{g_e}} \text { and } \mathrm{T}_m=2 \pi \sqrt{\frac{l}{g_m}} \\
\frac{\mathrm{T}_m}{\mathrm{~T}_e} & =\sqrt{\frac{g_e}{g_m}} \\
\mathrm{~T}_m & =\sqrt{\frac{g_e}{g_m}} \times \mathrm{T}_e=\sqrt{\frac{9.8}{1.7}} \times 3.5 \\
\mathrm{~T}_m & =8.4 \mathrm{sec}
\end{aligned}
$
Question 3.
A particle executes linear simple harmonic motion with an amplitude of $3 \mathrm{~cm}$. When the particle is at 2 $\mathrm{cm}$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then find its time period in seconds.

Answer:
$
\begin{aligned}
a=\omega^2 x, a=v, \mathrm{~A}=3 \mathrm{~cm}, x & =2 \mathrm{~cm} \\
v & =\omega \sqrt{\mathrm{A}^2-x^2} \\
\omega^2 x & =\omega \sqrt{\mathrm{A}^2-x^2} \\
\left(\frac{2 \pi}{\mathrm{T}}\right) 2 & =\sqrt{3^2-2^2} ; \frac{4 \pi}{\mathrm{T}}=\sqrt{5} \\
\mathrm{~T} & =\frac{4 \pi}{\sqrt{5}} \Rightarrow \mathrm{T}=5.6 \mathrm{sec}
\end{aligned}
$
Question 4.
A body of mass $m$ is attached to lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass $m$ is slightly pulled down and released, it oscillates with a time period of $3 \mathrm{~s}$. When the mass $\mathrm{m}$ is increased by $1 \mathrm{~kg}$, the time period of oscillations becomes $5 \mathrm{~s}$. Find the value of $\mathrm{m}$ is $\mathrm{kg}$.
Answer:
$
\begin{aligned}
& \mathrm{T}_1=3=2 \pi \sqrt{\frac{m}{k}} \\
& \mathrm{~T}_2=5=2 \pi \sqrt{\frac{m+1}{k}} \\
& \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{3}{5}=\sqrt{\frac{m}{m+1}} \\
& \frac{9}{25}=\frac{m}{m+1} \Rightarrow 9 m+9=25 m
\end{aligned}
$

$
m=\frac{9}{16} \mathrm{~kg}
$
Question 5.
Two simple harmonic motions are represented by the equations:
$
x_1=5 \sin \left(2 \pi t+\frac{\pi}{4}\right) ; x^2=5 \sqrt{2}(\sin 2 \pi t+\cos 2 \pi t)
$
What is the ratio of their amplitudes?
Answer:

$
\begin{aligned}
& x_1=5 \sin \left(2 \pi t+\frac{\pi}{4}\right) \quad\left[\therefore \mathrm{A}_1=5\right] \\
& x_2=5 \sqrt{2}\left(\sin 2 \pi t+\cos 2 \pi t^{\prime}\right. \\
& =10 \sin \left(\sin 2 \pi t \cos \frac{\pi}{4}+\cos 2 \pi t \sin \frac{\pi}{4}\right) \\
& x_2=10 \sin \left(2 \pi t+\frac{\pi}{4}\right) \quad\left[\therefore \mathrm{A}_2=10\right] \\
& \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{5}{10}=1: 2 \\
& A_1: A_2=1: 2 \\
&
\end{aligned}
$
Question 6.
A block whose mass is $1 \mathrm{~kg}$ is fastened to a spring. The spring has a spring constant of $50 \mathrm{Nm}^{-1}$. The block is pulled to a distance $\mathrm{x}=10 \mathrm{~cm}$ from its equilibrium position at $t=0$ on a frictionless surface from rest at $t=0$. Calculate the kinetic, potential and total energies of the block when it is $5 \mathrm{~cm}$ away from the mean position.
Answer:
Here $m=1 \mathrm{~kg}, k=50 \mathrm{Nm}^{-1}, \mathrm{~A}=10 \mathrm{~cm}=0.10 \mathrm{~m}, y=5 \mathrm{~cm}=0.05 \mathrm{~m}$
Kinetic energy: $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} k\left(\mathrm{~A}^2-y^2\right)=\frac{1}{2} \times 50\left[(0.10)^2-(0.05)^2\right]=0.1875 \mathrm{~J}$
Potential energy, $\mathrm{E}_{\mathrm{P}}=\frac{1}{2} k y^2=\frac{1}{2} \times 50 \times(0.05)^2=0.0625 \mathrm{~J}$
Total energy, $\quad \mathrm{E}=\mathrm{E}_{\mathrm{K}}+\mathrm{E}_{\mathrm{P}}=0.1875+0.0625=0.25 \mathrm{~J}$

Question 7.
A $5 \mathrm{~kg}$ collar is attached to a spring of force constant $500 \mathrm{Nm}^{-1}$. It slides without friction on a horizontal rod as shown in figure. The collar is displaced from its equilibrium position by $10.0 \mathrm{~cm}$ and released. Calculate:
(i) the period of oscillation
(ii) the maximum speed, and
(iii) the maximum acceleration of the collar.

Answer:
Here $m=5 \mathrm{~kg}, k=500 \mathrm{Nm}^{-1}, \mathrm{~A}=10.0 \mathrm{~cm}=0.10 \mathrm{~m}$
(i) Period of oscillation,
$
\mathrm{T}=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{5}{500}}=2 \times 3.14 \times \frac{1}{10} \mathrm{~s}=0.628 \mathrm{~s}
$
(ii) The maximum speed of the collar,
$
v_{\max }=\omega^2 \mathrm{~A}=\frac{k}{m} \mathrm{~A}=\frac{500}{5} \times 0.10=10 \mathrm{~ms}^{-2}
$
(iii) The maximum acceleration of the collar,
$
a_{\max }=\omega^2 \mathrm{~A}=\frac{k}{m} \mathrm{~A}=\frac{500}{5} \times 0.10=10 \mathrm{~ms}^{-2}
$
Question 8.
A $0.2 \mathrm{~kg}$ of mass hangs at the end of a spring. When $0.02 \mathrm{~kg}$ more mass is added to the end of the spring, it stretches $7 \mathrm{~cm}$ more. If the $0.02 \mathrm{~kg}$ mass is removed, what will be the period of vibration of the system?
Answer:
When $0.02 \mathrm{~kg}$ mass is added, the spring streches by $7 \mathrm{~cm}$
As
$
\begin{aligned}
& m g=k x \\
& \therefore \quad k=\frac{m g}{x}=\frac{0.02 \times 10}{7 \times 10^{-2}}=\frac{20}{7} \mathrm{Nm}^{-1} \\
&
\end{aligned}
$
When $0.02 \mathrm{~kg}$ mass ia removed, the period of ribration will be
$
\mathrm{T}=2 x \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{0.2}{20 / 7}}=2 \pi \sqrt{\frac{7}{100}}=\frac{2 \pi \times 2.645}{10}=1.66 \mathrm{~s}
$

Question 9.
A mass $\mathrm{M}$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period $\mathrm{T}$. If the mass is increased by $\mathrm{m}$, the time period becomes $5 \mathrm{~T} / 3$. What is the ratio $\mathrm{m} / \mathrm{M}$ ?
Answer: With mass $\mathrm{M}+m$, the time period becomes
$
\begin{aligned}
\frac{5 \mathrm{~T}}{3} & =2 \pi \sqrt{\frac{\mathrm{M}+m}{k}} \text { or } \frac{5}{3} \times 2 \pi \sqrt{\frac{\mathrm{M}}{k}}=2 \pi \sqrt{\frac{\mathrm{M}+m}{k}} \\
\frac{25}{9} \mathrm{M} & =\mathrm{M}+m \quad ; \frac{16}{9} \mathrm{M}=m \quad \text { or } \quad \frac{m}{\mathrm{M}}=\frac{16}{9}
\end{aligned}
$
Question 10.
A body describes simple harmonic motion with an amplitude of $5 \mathrm{~cm}$ and a period of $0.2 \mathrm{~s}$. Find the acceleration and velocity of the body when the displacement is $5 \mathrm{~cm}$.
Answer:

Here $\mathrm{A}=5 \mathrm{~cm}, \mathrm{~T}=0.2 \mathrm{~s}$
Velocity and acceleration at any displacement $\mathrm{x}$ are given by
$
\begin{aligned}
v & =\omega \sqrt{\mathrm{A}^2-x^2}=\frac{2 \pi}{\mathrm{T}} \sqrt{\mathrm{A}^2-x^2} \\
a & =-\omega^2 x=-\frac{4 \pi^2}{\mathrm{~T}^2} . . \\
\text { When } x=5 \mathrm{~cm} \quad a & =\frac{2 \pi}{0.2} \sqrt{5^2-5^2}=0 \\
a & =-\frac{4 \pi^2}{(0.2)^2} \times 5 \mathrm{cms}^{-2}=-500 \pi^2 \mathrm{~cm} \mathrm{~s}^{-2}=-5 \pi^2 \mathrm{~ms}^{-2}
\end{aligned}
$