Question 19: Oscillations

Question

$$ 	ext { Match List - I with List - II. } $$
$$
	ext { List - I }
$$
$$
	ext { List - II }
$$
A.
$$
\sin ^2 \omega t
$$
I.
Periodic with time period $T=\frac{\pi}{\omega}$ but not simple harmonic motion (SHM)
B.
$$
\sin ^3(2 \omega t)
$$
II.
Periodic with time period $T=\frac{2 \pi}{\omega}$ but Not SHM
C.
$$
\sin (\omega t)+\cos (\pi \omega t)
$$
III.
Periodic with time period $T=\frac{\pi}{\omega}$ and SHM
D.
$$
\cos \omega t+\cos 2 \omega t
$$
IV.
Non-periodic
Choose the correct answer from the options given below :

Accepted Answer

For the displacement equation $y=\sin ^2 \omega t$
Using the trigonometric identity $\sin ^2 	heta=\frac{1-\cos 2 	heta}{2}$. So, $\mathrm{y}=\frac{1}{2}-\frac{1}{2} \cos (2 \omega \mathrm{t})$.
For SHM, the acceleration must be proportional to displacement $\left(\frac{d^2 y}{d t^2}=-\omega^2 yight)$.
From the given equation, the function is shifted by $\frac{1}{2}$ and amplitude is $\frac{1}{2}$, so it oscillates between 0 and 1 .
While the oscillatory part is harmonic, the function represents a periodic motion and simple harmonic motion
The angular frequency is $2 \omega$.
$\mathrm{T}=\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}$.
Therefore, it is periodic with T = $\frac{\pi}{\omega}$ and SHM, i.e., $\mathrm{A} ightarrow \mathrm{III}$
For the displacement equation $y=\sin ^3(2 \omega t)$
Using the identity
$$ \sin 3 	heta=3 \sin 	heta-4 \sin ^3 	heta \Rightarrow \sin ^3 	heta=\frac{3 \sin 	heta-\sin 3 	heta}{4} $$
Here, $	heta=2 \omega \mathrm{t}$.
So, $\mathrm{y}=\frac{3 \sin (2 \omega \mathrm{t})-\sin (6 \omega \mathrm{t})}{4}=\frac{3}{4} \sin (2 \omega \mathrm{t})-\frac{1}{4} \sin (6 \omega \mathrm{t})$.
It is a superposition of two periodic functions $y_1=\frac{3}{4} \sin (2 \omega t)$ and $y_2=-\frac{1}{4} \sin (6 \omega t)$.
The time period is the LCM of time periods of both the periodic functions.
The period of first function is $\mathrm{T}_1=2 \pi / 2 \omega=\frac{\pi}{\omega}$.
The period of second function is $\mathrm{T}_2=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}$.
So, the time period of function is $T=\frac{\pi}{\omega}$.
As it contains multiple frequencies, so it is not SHM.
Therefore, it is a periodic function with time period $\mathrm{T}=\frac{\pi}{\omega}$ but not SHM. $(\mathrm{B} ightarrow \mathrm{I})$
For the displacement equation $y=\sin (\omega t)+\cos (\pi \omega t)$
It is a sum of two periodic functions $y_1=\sin (\omega t)$ and $y_2=\cos (\pi \omega t)$
It will be periodic only if the ratio of their time periods is a rational number.
$$ \omega_1=\omega \Rightarrow \mathrm{T}_1=\frac{2 \pi}{\omega} $$
$$ \begin{array}{ll}  & \omega_2=\pi \omega \Rightarrow \mathrm{T}_2=\frac{2 \pi}{\pi \omega}=\frac{2}{\omega} \ \operatorname{Ratio} \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\frac{2 \pi / \omega}{2 / \omega}=\pi & \end{array} $$
Since $\pi$ is an irrational number, the function is non-periodic. ( $\mathrm{C} ightarrow$ IV)
For the displacement equation $y=\cos \omega t+\cos 2 \omega t$
It is also the sum of two periodic function $y_1=\cos \omega t$ and $y_2=\cos 2 \omega t$
$$ \omega_1=\omega \Rightarrow \mathrm{T}_1=\frac{2 \pi}{\omega} 	ext { and } \omega_2=2 \omega \Rightarrow \mathrm{~T}_2=\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega} . $$
The ratio of time periods is, $\frac{\mathrm{T}_1}{\mathrm{~T}_2}=2$
The ratio is rational. The fundamental period is $\mathrm{T}=\frac{2 \pi}{\omega}$.
So, it is a periodic with time period $\mathrm{T}=\frac{\pi}{\omega}(\mathrm{D} ightarrow$ III $)$
Therefore, the correct match is A - II, B - I, C - IV, D - III.
Hence, the correct option is (D).

Answer

A-III, B-I, C-IV, D-II

Answer

A-II, B-I, C-III, D-IV

Answer

A-III, B-II, C-IV, D-I

Answer

A-II, B-I, C-IV, D-III

Oscillations

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