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Simple Harmonic Motion Problems




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 and comment at the end of the page.

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

Answer

(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 $32\frac{2^{1/2}}{\pi} \ m$

Answer

(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 x1 and x2, the corresponding velocities are u1 and u2. Show that time period of SHM is
$T=2\pi\left[\frac{x_2^2-x_1^2}{u_1^2-u_2^2}\right]^{1/2}$

Answer

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 ρ0. Density of the cylinder is ρ. If the cylinder is slightly depressed and released, then describe the motion of the cylinder

Answer

Mass of the cylinder (M) =SL ρ

In equilibrium position
Weight of the cylinder =Weight of Liquid displaced
Mg=Vρ0g
So V=M/ ρ0

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

Answer

(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)
$-A/2=\psi(0)=Asin{\varphi}$
$\varphi=\frac{\pi}{6}$



Question 6
A Bungee Jumper is attached to one end of a long elastic rope. The other end of the elastic rope is fixed to a high bridge. The Jumper steps off the bridge and falls from rest towards the river below. He does not hit the river below. The mass of the jumper is M and length of unstretched rope is L. Force constant of the rope is K. and gravitational field strength is g. Mass of rope is negligible ,air resistance is negligible.
1.Find out the distance y dropped by the jumper before coming instantaneously to rest for the first time
2.Maximum speed attained by the jumper during this drop
3.The time taken during the drop before coming to rest for the first time
Answer
y=[KL+mg+√(2mgKL+m2g2)]/k
v=√(2gL+mg2/k)
t=√(2L/g) + √(m/k)tan-1{-√(2KL/mg)}

Question 7
A mass attached to a spring is free to oscillate , with angular velocity ω , in a frictionless horizontal plane. The mass is displaced from it's equilibrium position by a distance x0 towards the center by pushing it with velocity v0 at time t=0. Find the amplitude of resulting oscillations in terms of ω,x0 and v0 . Answer
A=√[x02+(v02/◜)]

Question 8
A uniform cylinder of length l and mass m having cross-sectional area Ais suspended , with length vertical , from a fixed point by a massless spring , such that it is half submerged in a liquid of density σ at equilibrium position. When the cylinder is given a small downward push and released , it starts oscillating vertically with small amplitude . Calculate the frequency of oscillations of cylinder. Answer
f=[(k+(σAg)/m]1/2

Question 9
What should be the percentage change of length of pendulum in order that clock have same time period when moved from place where g=9.8 m/s2 to another where g=9.81 m/s2
Answer
.102%

Question 10
Two blocks of masses m1 and m2 are connected by a spring and these masses are free to oscillate along the axis of the spring. Find the angular frequency of oscillation.
Answer
ω=√[k(m1+m2)/m1m2]

Question 11
On a planet a freely falling body takes 2 sec when it is dropped from a height of 8 m, the time period of simple pendulum of length 1 m on that planet is

Answer

On a planet, if a body dropped initial velocity (u = 0) from a height h and takes time t to reach the ground
then $h=\frac {1}{2}g_p t^2$
or $g_p=4 \ m/s^2$
Using
$T=2\pi \sqrt {\frac {L}{g}}$
T=3.14 sec


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