Simple Harmonic Motion – JEE Main Previous Year Questions with Solutions

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Previous Years AIEEE/JEE Mains Questions

Q. If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time ?

$(1) \mathrm{aT}+2 \pi \mathrm{v}$

(2) $\frac{\mathrm{aT}}{\mathrm{v}}$

(3) $a^{2} \mathrm{T}^{2}+4 \pi^{2} \mathrm{v}^{2}$

(4) $\frac{\mathrm{aT}}{\mathrm{x}}$

[AIEEE-2009]

Sol. (3,4)

\frac{\mathrm{aT}}{\mathrm{x}}=\frac{\omega^{2} \mathrm{xT}}{\mathrm{x}}=\frac{4 \pi^{2}}{\mathrm{T}}=\mathrm{constant}

\mathrm{a}^{2} \mathrm{T}^{2}+4 \pi^{2} \mathrm{v}^{2}

=\left(\omega^{2} \mathrm{x}\right)^{2} \mathrm{T}^{2}+4 \pi^{2} \omega^{2}\left(\mathrm{A}^{2}-\mathrm{x}^{2}\right)

=\left(\frac{4 \pi}{\mathrm{T}^{2}}\right)^{2} \mathrm{x}^{2} \cdot \mathrm{T}^{2}+4 \pi^{2} \cdot \frac{4 \pi^{2}}{\mathrm{T}^{2}} \cdot\left(\mathrm{A}^{2}-\mathrm{x}^{2}\right)

$=\frac{16 \pi^{2}}{\mathrm{T}^{2}} \mathrm{A}^{2}=$ constant


Q. A mass M, attached to a horizontal spring, executes S.H.M. with amplitude $\mathrm{A}_{1}$. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude $\mathrm{A}_{2}$. The ratio of $\left(\frac{\mathrm{A}_{1}}{\mathrm{A}_{2}}\right)$ is :-

( 1)$\left(\frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}\right)^{1 / 2}$

( 2)$\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}\right)^{1 / 2}$

(3) $\frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}$

(4) $\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}$

[AIEEE-2011]

Sol. (2)

Energy of simple harmonic oscillator is constant.

$\Rightarrow \frac{1}{2} \mathrm{M} \omega^{2} \mathrm{A}_{1}^{2}=\frac{1}{2}(\mathrm{m}+\mathrm{M}) \omega^{2} \mathrm{A}_{2}^{2}$

$\frac{\mathrm{A}_{1}^{2}}{\mathrm{A}_{2}^{2}}=\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}$

$\therefore \frac{\mathrm{A}_{1}}{\mathrm{A}_{2}}=\sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}}$


Q. Two particles are executing simple harmonic motion of the same amplitude A and frequency  along the x-axis. Their mean position is separated by distance $\mathrm{X}_{0}\left(\mathrm{X}_{0}>\mathrm{A}\right)$. If the maximum separation between them is $\left(\mathrm{X}_{0}+\mathrm{A}\right)$, the phase difference between their motion is :-

(1) $\frac{\pi}{4}$

(2) $\frac{\pi}{6}$

(3) $\frac{\pi}{2}$

(4) $\frac{\pi}{3}$

[AIEEE-2011]

Sol. (4)

$\mathrm{x}_{1}=\mathrm{A} \sin \left(\omega \mathrm{t}+\phi_{1}\right)$

$\mathrm{x}_{2}=\mathrm{A} \sin \left(\omega \mathrm{t}+\phi_{2}\right)$

$\mathrm{x}_{1}-\mathrm{x}_{2}=\mathrm{A}\left[2 \sin \left[\omega \mathrm{t}+\frac{\phi_{1}+\phi_{2}}{2}\right] \sin \left[\frac{\phi_{1}-\phi_{2}}{2}\right]\right]$

$\mathrm{A}=2 \mathrm{A} \sin \left(\frac{\phi_{1}-\phi_{2}}{2}\right)$

$\frac{\phi_{1}-\phi_{2}}{2}=\frac{\pi}{6}$

$\phi_{1}=\frac{\pi}{3}$


Q. If a spring of stiffness’ $\mathrm{k}^{\prime}$ ‘is cut into two parts’ ‘A’ and ‘B’ of length $\ell_{\mathrm{A}}: \ell_{\mathrm{B}}=2: 3,$ then the stiffness

of spring ‘A’ is given by :-

(1) $\frac{5}{2} \mathrm{k}$

(2) $\frac{3 k}{5}$

(3) $\frac{2 \mathrm{k}}{5}$

(4) k

[AIEEE-2011]

Sol. (1)

We know that,

$\mathrm{k} \propto \frac{1}{\ell_{0}}$

$\therefore \mathrm{k}_{\mathrm{A}}=\frac{5}{2} \mathrm{k}$


Q. If a simple pendulum has Significant amplitude (up to a factor of 1/ e of original) only in the period between $\mathrm{t}=0 \mathrm{s}$ to $\mathrm{t}=\tau \mathrm{s}$, then $\tau$ may be called the average life ofthe pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with ‘b’ as the constant of proportionality, the average lifetime ofthe pendulum is (assuming damping is small) in seconds:

(1) $\frac{1}{b}$

(2) $\frac{2}{\mathrm{b}}$

(3) $\frac{0.693}{\mathrm{b}}$

(4) b

[AIEEE-2012]

Sol. (2)

Average life $=\tau=\frac{2}{\mathrm{b}}$


Q. The amplitude of adamped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to a times its original magnitude, where a equals

(1) 0.81            (2) 0.729               (3) 0.6                  (4) 0.7

[JEE Main-2013]

Sol. (2)

$\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-\mathrm{t} / \mathrm{e}} \Rightarrow \frac{\mathrm{A}_{2}}{\mathrm{A}_{1}}=\frac{\mathrm{e}^{-\mathrm{t}_{2} / \mathrm{e}}}{\mathrm{e}^{-\mathrm{t} / \mathrm{e}}}=\frac{\left(\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{\tau}\right)}{\mathrm{e}}=\frac{-10}{\mathrm{c}}$

Also $0.9=\mathrm{A}_{0} \mathrm{e}^{-5 / \mathrm{c}} \Rightarrow \mathrm{e}^{-5 / \mathrm{c}}=0.9$

so $\mathrm{A}_{2}=\left(0.9 \mathrm{A}_{0}\right)(0.9)^{2}=0.729 \mathrm{A}_{0}$


Q. A particle moves with simple harmonic motion in a straight line. In first $\tau$ s, after starting from

rest it travels a distance a, and in next $\tau$ s it travels $2 a,$ in same direction, then :

(1) Amplitude of motion is 4a

(2) Time period of oscillation is $6 \tau$

(3) Amplitude of motion is 3a

(4) Time period of oscillation is $8 \tau$

[JEE Mains-2014]

Sol. (4)


Q. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly ? (graphs are schematic and not drawn to scale)

[JEE Mains-2015]

Sol. (4)

$\mathrm{KE}=\frac{1}{2} \mathrm{m} \omega^{2}\left[\mathrm{A}^{2}-\mathrm{d}^{2}\right],$ downward parabola


Q. A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance $\frac{2 \mathrm{A}}{3}$ from equilibrium position. The new amplitude of the motion is :-

(1) $\frac{7 \mathrm{A}}{3}$

(2) $\frac{\mathrm{A}}{3} \sqrt{41}$

(3) 3A

(4) A $\sqrt{3}$

[JEE Mains-2016]

Sol. (1)

Let new amplitude is A’

initial velocity

$\mathrm{v}^{2}=\omega^{2}\left(\mathrm{A}^{2}-\left(\frac{2 \mathrm{A}}{3}\right)^{2}\right) \quad \ldots(1)$

Where $\mathrm{A}$ is initial amplitude $\& \omega$ is angular frequency.

Final velocity

$(3 \mathrm{v})^{2}=\omega^{2}\left(\mathrm{A}^{\prime 2}-\left(\frac{2 \mathrm{A}}{3}\right)^{2}\right) \quad \ldots(2)$

From equation & equation (2)

$\frac{1}{9}=\frac{A^{2}-\frac{4 A^{2}}{9}}{A^{\prime 2}-\frac{4 A^{2}}{9}}$

$\mathrm{A}^{\prime}=\frac{7 \mathrm{A}}{3}$


Q. A particle is executing simple harmonic motion with a time period T. AT time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like

[JEE Mains-2017]

Sol. (2)

Time taken to reach the extreme position from equilibrium position is $\frac{\mathrm{T}}{4}$. Velocity is maximum at equilibrium position and zero at extreme position.

$\mathrm{V}=\mathrm{A} \omega \cos \omega \mathrm{t}$

$\mathrm{K.E.}=\frac{1}{2} \mathrm{mv}^{2}$

(m is the mass of particle and v is the velocity of particle

$\mathrm{K.E.}=\frac{1}{2} \mathrm{mA}^{2} \omega^{2} \cos ^{2} \omega \mathrm{t}$

Hence graph of K.E. v/s time is square cos function


Q. frequency of $10^{12} / \mathrm{sec} .$ What is the force constant of the bonds connecting one atom with

the other ? (Mole wt. of silver $=108$ and Avagadro number $=6.02 \times 10^{23} \mathrm{gm}$ mole $^{-1}$ )

(1) 7.1 N/m                (2) 2.2 N/m                 (3) 5.5 N/m              (4) 6.4 N/m

[JEE Mains-2018]

Sol. (1)

Time period of SHM is given by

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

frequency $=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}=10^{12}$

where m = mass of one atom

$=\frac{108}{\left(6.02 \times 10^{23}\right)} \times 10^{-3} \mathrm{kg}$

$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{108 \times 10^{-3}} \times 6.02 \times 10^{23}}=10^{12}$

On solving K = 7.1 N/m


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