NEET Oscillation MCQs & PYQs

Practice NEET Oscillation Questions

Question 121:

moderate

A particle is executing S.H.M. along a straight line. Its velocities at distances $ x_1 $ and $ x_2 $ from the mean position are $ v_1 $ and $ v_2 $, respectively. Its time period is:

(2015)

Velocity in SHM is $ v^2 = \omega^2(A^2 - x^2) $. So, $ v_1^2 = \omega^2(A^2 - x_1^2) $ and $ v_2^2 = \omega^2(A^2 - x_2^2) $. Subtracting these equations gives $ v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2) $, yielding $$ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} $$.

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Question 122:

easy

A particle is executing a simple harmonic motion. Its maximum acceleration is $ \alpha $ and maximum velocity is $ \beta $. Then, its time period of vibration will be:

(2015 Re)

Maximum acceleration is $ a_{\text{max}} = A\omega^2 = \alpha $ and maximum velocity is $ v_{\text{max}} = A\omega = \beta $. Dividing them, we get $ \omega = \frac{\alpha}{\beta} $. Thus, the time period is $ T = \frac{2\pi}{\omega} = \frac{2\pi \beta}{\alpha} $.

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Question 123:

easy

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:

(2020)

Displacement is given by $ y = A \sin(\omega t) $ and acceleration is $ a = -A\omega^2 \sin(\omega t) = A\omega^2 \sin(\omega t + \pi) $. Thus, the phase difference is $ \pi \text{ rad} $.

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Question 124:

easy

Identify the function which represents a periodic motion.

(2020-Covid)

The function $ \sin \omega t + \cos \omega t $ represents a superposition of two simple harmonic motions, making it periodic. The other given functions do not repeat their values over equal intervals of time.

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Question 125:

easy

Average velocity of a particle executing SHM in one complete vibration is :

(2019)

In one complete vibration, the particle returns to its starting point, so the net displacement is zero. Since average velocity is total displacement divided by total time, it is zero.

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Question 126:

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)

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 127:

easy

Out of the following functions representing motion of a particle which represents S.H.M.:

(i) $y = \sin \omega t – \cos \omega t$

(ii) $y = \sin^3 \omega t$

(iii) $y = 5\cos\left(\frac{3\pi}{4} – 3\omega t\right)$

(iv) $y = 1 + \omega t + \omega^2 t^2$

(2011 Pre)

(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 128:

easy

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)

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 129:

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)

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 130:

easy

Which one of the following equations of motion represents simple harmonic motion?

where $k$, $k_0$, $k_1$ and $a$ are all positive.

(2009)

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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