Oscillation - NEET Physics Questions
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Oscillation

Practice Questions:
Question 11: moderate

Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to paths of the two particles. The phase difference is: (2011 Mains)

1. $0$
2. $\frac{2\pi}{3}$
3. $\pi$
4. $\frac{\pi}{6}$
View Answer

Let displacement be $x = A\sin(\omega t + \phi)$. When they cross, $x = A/2$. $A/2 = A\sin(\phi) \Rightarrow \sin(\phi) = 1/2$. The two phases are $\pi/6$ and $5\pi/6$ (since moving in opposite directions). Phase difference $= 5\pi/6 - \pi/6 = 4\pi/6 = 2\pi/3$.

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Question 12: moderate

Out of the following functions representing motion of a particle which represents S.H.M.: (2011 Pre)n(i) $y = \sin \omega t – \cos \omega t$n(ii) $y = \sin^3 \omega t$n(iii) $y = 5\cos\left(\frac{3\pi}{4} – 3\omega t\right)$n(iv) $y = 1 + \omega t + \omega^2 t^2$

1. Only (i)
2. Only (iv) does not represent SHM
3. Only (i) and (iii)
4. Only (i) and (ii)
View Answer

(i) Linear combination of sine and cosine represents SHM. (ii) $y = \sin^3\omega t$ is an oscillatory motion but not SHM. (iii) Simple cosine function with phase shift represents SHM. (iv) Non-oscillatory. Thus, only (i) and (iii) represent SHM.

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Question 13: moderate

A particle moves in a circle of radius $5 \text{ cm}$ with constant speed and time period $0.2\pi$. The acceleration of the particle is: (2011 Pre)

1. $15 \text{ m/s}^2$
2. $25 \text{ m/s}^2$
3. $36 \text{ m/s}^2$
4. $5 \text{ m/s}^2$
View Answer

Radius $r = 5 \text{ cm} = 0.05 \text{ m}$, Time period $T = 0.2\pi$. Angular velocity $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2\pi} = 10 \text{ rad/s}$. Centripetal acceleration $a = \omega^2 r = (10)^2 \times 0.05 = 100 \times 0.05 = 5 \text{ m/s}^2$.

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Question 14: moderate

The displacement of a particle along the x-axis is given by $x = a\sin^2\omega t$. The motion of the particle corresponds to: (2010 Pre)

1. Simple harmonic motion of frequency $\omega/2\pi$
2. Simple harmonic motion of frequency $\omega/\pi$
3. Simple harmonic motion of frequency $3\omega/2\pi$
4. Non simple harmonic motion
View Answer

Equation is $x = a\sin^2\omega t = \frac{a}{2}(1 - \cos 2\omega t)$. This represents SHM about the mean position $x = a/2$. The angular frequency is $2\omega$. The frequency is $f = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}$.

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Question 15: moderate

Which one of the following equations of motion represents simple harmonic motion? (2009)nwhere $k$, $k_0$, $k_1$ and $a$ are all positive.

1. Acceleration $= -k(x)$
2. Acceleration $= k(x + a)$
3. Acceleration $= kx$
4. Acceleration $= -k_0x + k_1x^2$
View Answer

For simple harmonic motion, the acceleration must be directly proportional and opposite in direction to the displacement. Thus, $a \propto -x$, which matches the equation $\text{Acceleration} = -k(x)$.

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Question 16: moderate

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $a$ and time period $T$. The speed of pendulum at $x = a/2$ will be: (2009)

1. $\frac{\pi a}{T}$
2. $\frac{3\pi^2 a}{T}$
3. $\frac{\pi a \sqrt{3}}{T}$
4. $\frac{\pi a \sqrt{3}}{2T}$
View Answer

Velocity in SHM is given by $v = \omega\sqrt{A^2 - x^2}$. Here $A = a$, $x = a/2$, and $\omega = \frac{2\pi}{T}$. So, $v = \frac{2\pi}{T}\sqrt{a^2 - \frac{a^2}{4}} = \frac{2\pi}{T} \times \frac{a\sqrt{3}}{2} = \frac{\pi a \sqrt{3}}{T}$.

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Question 17: moderate

A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = a \sin(\omega t + \pi/6)$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity? (2008)

1. $\frac{T}{12}$
2. $\frac{T}{8}$
3. $\frac{T}{6}$
4. $\frac{T}{3}$
View Answer

Velocity $v = \frac{dx}{dt} = a\omega\cos(\omega t + \pi/6)$. Maximum velocity is $a\omega$. Given $v = \frac{a\omega}{2}$, so $\cos(\omega t + \pi/6) = 1/2$. This gives $\omega t + \pi/6 = \pi/3 \Rightarrow \omega t = \pi/6$. Since $\omega = 2\pi/T$, we get $\frac{2\pi t}{T} = \frac{\pi}{6} \Rightarrow t = \frac{T}{12}$.

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Question 18: moderate

Two Simple Harmonic Motions of angular frequency $100 \text{ rad s}^{-1}$ and $1000 \text{ rad s}^{-1}$ have the same displacement amplitude. The ratio of their maximum accelerations is: (2008)

1. $1 : 10^4$
2. $1 : 10$
3. $1 : 10^2$
4. $1 : 10^3$
View Answer

Maximum acceleration in SHM is given by $a_{\text{max}} = \omega^2 A$. Since amplitude $A$ is the same, $a_{\text{max}} \propto \omega^2$. The ratio is $a_1 / a_2 = (\omega_1 / \omega_2)^2 = (100 / 1000)^2 = (1/10)^2 = 1 : 100 = 1 : 10^2$.

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Question 19: moderate

The circular motion of a particle with constant speed is: (2005)

1. Simple harmonic but not periodic
2. Periodic and simple harmonic
3. Neither periodic nor simple harmonic
4. Periodic but not simple harmonic
View Answer

Uniform circular motion repeats itself after fixed intervals of time, making it periodic.\nHowever, the motion itself is not along a straight line towards a mean position, so it is not simple harmonic.

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Question 20: moderate

Which one of the following statements is true for the speed ‘$v$’ and the acceleration ‘$a$’ of a particle executing simple harmonic motion? (2004)

1. Value of '$a$' is zero, whatever may be the value of '$v$'
2. When '$v$' is zero, '$a$' is zero
3. When '$v$' is maximum, '$a$' is zero
4. When '$v$' is maximum, '$a$' is maximum
View Answer

In simple harmonic motion, speed is maximum at the mean position.\nAt this mean position, the displacement is zero, causing the restoring force and acceleration to be zero.

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