Numerical Problems-1 - Chapter 10 - Oscillations - 11th Science Guide Samacheer Kalvi Solutions

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Numerical Problems
Question 1.
Consider the Earth as a homogeneous sphere of radius $\mathrm{R}$ and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time period is
$
T=2 \pi \sqrt{\frac{R}{g}}
$

Answer:
Oscillations of a particle dropped in a tunnel along the diameter of the earth. Consider earth to be a sphere of radius $\mathrm{R}$ and centre $O$. A straight tunnel is dug along the diameter of the
earth. Let ' $g$ ' be the value of acceleration due to gravity at the surface of the earth. Suppose a body of mass ' $\mathrm{m}$ ' is dropped into the tunnel and it is at point $R$ i.e., at a depth $d$ below the surface of the earth at any instant.
If $g^{\prime}$ is acceleration due to gravity at $P$. then
$
g^{\prime}=g\left(1-\frac{d}{\mathrm{R}}\right)=g\left(\frac{\mathrm{R}-d}{\mathrm{R}}\right)
$
If $y$ is distance of the body from the centre of the earth, then
$
\begin{aligned}
\mathrm{R}-d & =y \\
\therefore \quad g^{\prime} & =g\left(\frac{y}{\mathrm{R}}\right)
\end{aligned}
$
Force acting on the body a point $P$ is
$
\mathrm{F}=-m g^{\prime}=-\frac{m g}{\mathrm{R}} y \text { i.e., } \mathrm{F} \propto y
$
Negative sign indicates that the force acts in the opposite direction of displacement.
Thus the body will execute SHM with force constant, $k=\frac{m g}{\mathrm{R}}$
The period of oscillation of the body will be $\mathrm{T}=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{m g / \mathrm{R}}}$
$
T=2 \pi \sqrt{\frac{R}{g}}
$
Question 2.
Calculate the time period of the oscillation of a particle of mass $\mathrm{m}$ moving in the potential defined as

$\mathrm{U}(x)=\left\{\begin{array}{l}
\frac{1}{2} k x^2, x<0 \\
m g x, x>0
\end{array}\right.$

Answer:
$
\mathrm{U}=\frac{1}{2} k x^2, x<0
$
Force from potential energy, $\mathrm{F}=-\frac{d \mathrm{U}}{d x}=\frac{-d}{d x}\left[\frac{1}{2} k x^2\right]$
$
=\frac{-1}{2} k \frac{d}{d x}\left(x^2\right)=\frac{-1}{2} k(2 x) ; \mathrm{F}=-k x
$
According to newton's 2nd law, $\mathrm{F}=m a$
$
\begin{array}{r}
m a \doteq-k x \\
a=-\frac{k x}{m}
\end{array}
$
Acceleration of particle in SHM,
$
\begin{aligned}
(1)=(2) \Rightarrow \frac{-k x}{m}=-\omega^2 x & =-\omega^2 x \\
\omega^2 & =\frac{k}{m} \Rightarrow \omega=\sqrt{\frac{k}{m}} \\
\text { Time period of the oscillation, } \mathrm{T} & =\frac{2 \pi}{\omega} \\
\mathrm{T} & =2 \pi \sqrt{\frac{m}{k}}
\end{aligned}
$
Question 3.
Consider a simple pendulum of length $1=0.9 \mathrm{~m}$ which is properly placed on a trolley rolling down on a inclined plane which is at $\theta=45^{\circ}$ with the horizontal. Assuming that the inclined plane is frictionless, calculate the time period of oscillation of the simple pendulum.
Answer:
Length of the pendulum $1=0.9 \mathrm{~m}$
Inclined angle $\theta=45^{\circ}$
Time period of a simple pendulum $\mathrm{T}=2 \pi \sqrt{\frac{l}{g^{\prime}}}$
$
\begin{aligned}
& g^{\prime}=g \cos \theta \\
& \mathrm{T}=2 \pi \sqrt{\frac{l}{g \cos \theta}}=2 \times 3.14 \sqrt{\frac{0.9}{9.8 \times \cos 45^{\circ}}}=6.28 \times \sqrt{0.1298} \\
& \mathrm{~T}=2.263 \mathrm{~s}
\end{aligned}
$

Question 4.
A piece of wood of mass $m$ is floating erect in a liquid whose density is $p$. If it is slightly pressed down and released, then executes simple harmonic motion. Show that its time period of oscillation is
$
T=2 \pi \sqrt{\frac{m}{\mathrm{Ag} \rho}}
$
Answer:
Spring factor of liquid $(k)=\operatorname{A\rho g}$
Inertia factor of piece of wood $=m$
$
\text { Time period } \mathrm{T}=2 \pi \sqrt{\frac{\text { Intertia factor }}{\text { Spring factor }}} ; T=2 \pi \sqrt{\frac{m}{A \rho g}}
$
Question 5.
Consider two simple harmonic motion along $\mathrm{x}$ and $\mathrm{y}$-axis having same frequencies but different amplitudes as $\mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\varphi)$ (along $\mathrm{x}$ axis) and $\mathrm{y}=\mathrm{B} \sin \omega t$ (along $\mathrm{y}$ axis).
Then show that $\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}-\frac{2 x y}{\mathrm{AB}} \cos \phi=\sin ^2 \phi$ and also discuss the special cases when
(a) $\phi=0$
(b) $\phi=\pi$
(c) $\phi=\frac{\pi}{2}$
$(d) \phi=\frac{\pi}{2}$ and $\mathrm{A}=\mathrm{B}$
(e) $\phi=\frac{\pi}{4}$
Note: when a particle is subjected to two simple harmonic motion at right angle to each other the particle may move along different paths. Such paths are called Lissajous figures.
Answer:
(a) $y=\frac{\mathrm{B}}{\mathrm{A}} x$, equation is a straight line passing through origin with positive slope.
(b) $y=-\frac{B}{A} x$, equation is a straight line passing through origin with negative slope.
(c) $\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1$, equation is an ellipse whose center is origin. A2 B2
(d) $x^2+y^2=\mathrm{A}^2$, equation is a circle whose center is origin.
(e) $\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}-\frac{2 x y}{\mathrm{AB}} \frac{1}{\sqrt{2}}=\frac{1}{2}$, equation is an ellipse which (oblique ellipse which means tilted ellipse)

Question 6.
Show that for a particle executing simple harmonic motion
(a) the average value of kinetic energy is equal to the average value of potential energy.
(b) average potential energy $=$ average kinetic energy $=\frac{1}{2}$ (total energy)
[Hint: average kinetic energy $=<$ kinetic energy $>=\frac{1}{\mathrm{~T}} \int_0^{\mathrm{T}}$ (Kinetic energy) $d t$ and average Potential energy $=<$ Potential energy $>=\frac{1}{T} \int_0^T($ Potential energy $\left.) d t\right]$
Suppose a particle of mass $m$ executes SHM of period $T$. The displacement of the particles at any instant $t$ is given by $y=A \sin \omega t$

Answer:

Velocity $v=\frac{d y}{d t}=\omega \mathrm{A} \cos \omega t$
Kinetic energy $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} m v^2=\frac{1}{2} m \omega^2 \mathrm{~A}^2 \cos ^2 \omega t$
Potential energy, $\mathrm{E}_{\mathrm{P}}=\frac{1}{2} m \omega^2 y^2=\frac{1}{2} m \omega^2 \mathrm{~A}^2 \sin ^2 \omega t$
(a) Average K.E. over a period of oscillation,
$
\begin{aligned}
\mathrm{E}_{\mathrm{K}_{a v}} & =\frac{1}{\mathrm{~T}} \int_0^{\mathrm{T}} \mathrm{E}_{\mathrm{K}} d t=\frac{1}{\mathrm{~T}} \int_0^{\mathrm{T}} \frac{1}{2} m \omega^2 \mathrm{~A}^2 \cos ^2 \omega t d t \\
& =\frac{1}{2 \mathrm{~T}} m \omega^2 \mathrm{~A}^2 \int_0^{\mathrm{T}}\left(\frac{1+\cos 2 \omega t}{2}\right) d t \\
& =\frac{1}{4 \mathrm{~T}} m \omega^2 \mathrm{~A}^2\left[t+\frac{\sin 2 \omega t}{2 \omega}\right]_0^{\mathrm{T}}=\frac{1}{4 \mathrm{~T}} m \omega^2 \mathrm{~A}^2 \mathrm{~T} \\
\mathrm{E}_{\mathrm{K}_{o v}} & =\frac{1}{4} m \omega^2 \mathrm{~A}^2
\end{aligned}
$
(b) Average P.E. over a period of oscillation
$
\begin{aligned}
& \mathrm{E}_{\mathrm{P}_{a v}}=\frac{1}{\mathrm{~T}} \int_0^{\mathrm{T}} \mathrm{E}_{\mathrm{P}} d t=\frac{1}{\mathrm{~T}} \int_0^{\mathrm{T}} \frac{1}{2} m \omega^2 \mathrm{~A}^2 \sin ^2 \omega t d t \\
&=\frac{1}{2 \mathrm{~T}} m \omega^2 \mathrm{~A}^2 \int_0^{\mathrm{T}}\left(\frac{1-\cos 2 \omega t}{2}\right) d t \\
&=\frac{1}{4 \mathrm{~T}} m \omega^2 \mathrm{~A}^2\left[t-\frac{\sin 2 \omega t}{2 \omega}\right]_0^{\mathrm{T}}=\frac{1}{4 \mathrm{~T}} m \omega^2 \mathrm{~A}^2 \mathrm{~T} \\
& \mathrm{E}_{\mathrm{P}_{\mathrm{av}}}=\frac{1}{4} m \omega^2 \mathrm{~A}^2
\end{aligned}
$
Question 7.
Compute the time period for the following system if the block of mass $m$ is slightly displaced vertically down from Its equilibrium position and then released. Assume that the pulley is light and smooth, strings and springs are light.
Answer:

Case (a): Pulley is fixed rigidly here. When the mass displace by y and the spring will also stretch by y.
Therefore, $F=T=\mathrm{ky}$
$
\mathrm{T}=2 \pi \sqrt{\frac{m}{k}}
$
Case (b): Mass displace by y, pulley also displaces by y. $T=4 \mathrm{ky}$
$
\mathrm{T}=2 \pi \sqrt{\frac{m}{4 k}}
$