In this page we have *Simple Harmonic Motion Problems for JEE Main/Advanced* . Hope you like them and do not forget to like , social share
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## Subjective Questions

**Question 1**
A particle is situated at the end of the one arm of a tuning fork passes through its equilibrium position with a velocity of 2 ms

^{-1}.The amplitude is 10

^{-3} m

(a) Determine the frequency and time period of the tuning fork

(b) Write the equations expressing its displacement and velocity as a function of time .Assume initial phase is zero

Solution
(a) We know that when any particle passes through its equilibrium position its velocity is maximum .Thus

$v_{max}=2 \ m/s$

In SHM $v_{max}=\pm\omega A$

Where $\omega$ is the angular frequency of SHM

Now given that $A=10^{-3} \ m$

$\omega =2/10^{-3} =2000 \ rad/sec$

Now frequency of oscillations

$f=\frac{\omega}{2\pi}$

=318.3 Hz

Again, Time period

$T=1/f= \pi \times 10^{-3} \ s$

(b)

Displacement of particle executing SHM is given by

$x=Acos{(}\omega t+\varphi)$

Given that initial phase $\varphi=0$ and $A=10^{-3} \ m$

Thus

$x=10^{-3}cos{(}2000t)$

Again, velocity of SHM is given by

$v=A\omega sin{\omega}t$

Putting all the values

$v=-2sin{(}2000t)$

**Question 2**
A 4 kg particle is moving along the x-axis under the action of the force

$F=-\left(\frac{\pi^2}{16}\right)x \ N$
When t=2 sec the particle passes through the origin and when t=4 sec.its velocity is 4 m/s

(a) Find the equation of the displacement

(b) Show that the amplitude of motion is $32\frac{2^{1/2}}{\pi} \ m$

Solution
(a) Given that

$F=-\left(\frac{\pi^2}{16}\right)x $

We know that for SHM, force is given by

$F=-kx$

Where k is the force constant

Comparing SHM general equation with the given force equation

$k=\frac{\pi^2}{16}$

Now angular velocity in SHM is given by

$\omega=\sqrt{\frac{k}{m}}$

=> $\omega=\sqrt{\frac{\pi^2}{16m}}$

Now m=4 kg

$ \omega =\pi /8$

Now at t=2sec,x=0 (equilibrium position) we know that displacement of particle is SHM is given by

$x=Acos{(}\omega t+\varphi)$

Now $ \omega =\pi /8$

At t=2 x=0

$0=Acos{(}\frac{\pi}{8}*2+\varphi)$

Or $\varphi= \pi /4$

So equation of displacement

$x=Acos{(}\frac{\pi}{8}t+\frac{\pi}{4})$

Where A is the amplitude of SHM

(b) Velocity of SHM is

$v=-A\omega s i n{(}\omega t+\varphi)$

In this case,

$v=-A\frac{\pi}{8}sin{(}\frac{\pi}{8}t+\frac{\pi}{4})$

Given that t=4 sec v=4 m/sec

$4=-A\frac{\pi}{8}sin{(}\frac{\pi}{8}4+\frac{\pi}{4})$

$A=-32\frac{2^{1/2}}{\pi}m$

Here negative sign indicates that particle was moving with -x direction at 4 sec

**Question 3**
A particle is moving with SHM in a straight line, when the distance of the particle from the equilibrium position has the values x

_{1} and x

_{2,} the corresponding velocities are u

_{1} and u

_{2}. Show that time period of SHM is

$T=2\pi\left[\frac{x_2^2-x_1^2}{u_1^2-u_2^2}\right]^{1/2}$

Solution
Let O be the equilibrium position of particle. If P and Q are the two positions of the particles at distance $x_1$ and $x_2$ from O respectively, the velocity $u_1$ at the position P is given by

$u_1=\omega\sqrt{A^2-x_1^2}$ ---(1)

The velocity $u_2$ at position Q is given by

$u_2=\omega\sqrt{A^2-x_2^2}$ ----(2)

Squaring 1 and 2 and subtracting

$u_1^2-u_2^2=\omega^2(x_2^2-x_1^2)$

Or

$\omega=\sqrt{\frac{u_1^2-u_2^2}{x_2^2-x_1^2}}$

Hence the Time period

$T=2\pi\left[\frac{x_2^2-x_1^2}{u_1^2-u_2^2}\right]^{1/2}$

**Question 4**
A cylinder of cross-sectional area S and length L is floating in a liquid of density ρ

_{0}. Density of the cylinder is ρ. If the cylinder is slightly depressed and released, then describe the motion of the cylinder

Solution
Mass of the cylinder (M) =SL ρ

In equilibrium position

Weight of the cylinder =Weight of Liquid displaced

Mg=Vρ_{0}g

So V=M/ ρ_{0}

Let the cylinder is displaced by x from the mean position, then it will displace additional liquid of volume Sx.

So upward force action on the cylinder

$F= -Sx \rho _0g$

$Ma=-Sx \rho _0g$

$a=-(\frac{S\rho_0g}{M})x$

$a=-(\frac{S\rho_0g}{SL\rho})x$

$a=-\left(\frac{\rho_0g}{\rho L}\right)x$

Since a=-kx

So the motion is SHM

Time period of SHM

$T=2\pi\sqrt{\frac{\rho L}{\rho_0g}}$

**Question 5**
A particle executes SHM with amplitude A.If its starting point from rest is

(a) ψ =+A

(b) ψ =-A

(c) ψ =+A/2

Find the different values of the phase constant (Φ) for the equations

$\psi(t)=Acos{(}\omega t+\varphi)$

$\psi(t)=Asin{(}\omega t+\varphi)$

Solution
(a) t=0 , ψ =+A

Equation (A)

$\psi(0)=Acos{(}\omega*0+\varphi)$

$\varphi=0$

Equation (B)

$\psi(0)=Asin{(}\omega*0+\varphi)$

(b) t=0 , ψ =-A

Equation (A)

$-A=\psi(0)=Acos{\varphi}$

$\varphi=\pi$

Equation (B)

$-A=\psi(0)=Asin{\varphi}$

$\varphi=\frac{3\pi}{2}$

(c) t=0, ψ =+A/2

Equation (A)

$-A/2=\psi(0)=Acos{\varphi}$

$\varphi=\pi/3$

Equation (B)

$-A/2=\psi(0)=Asin{\varphi}$

$\varphi=\frac{\pi}{6}$

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