In this page we have Important Objective type questions on Simple Harmonic Motion for JEE main/Advanced . Hope you like them and do not forget to like , social share
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Linked type Comprehensions
(A)A body of mass 36 g moves with SHM of amplitude A=13 cm and period T=12s
At t=0 x=+13 cm
Question 1
Find the velocity when x=5 cm
(a) $\pm6.28 \ cm/sec$
(b) $\pm6.00 \ cm/sec$
(c) $\pm7.28 \ cm/sec$
(d) $\pm5.28 \ cm/sec$
Solution
We know that for SHM
$v=\pm\omega\sqrt{\left(A^2-x^2\right)}$
Where A is amplitude and $\omega=2 \pi /T$
Solving we get
$v=\pm6.28cm/sec$
Question 2
Find the
displacement at t=2 sec
(a) 7.0 cm
(b) 6.5 cm
(c) 6 cm
(d) None of these
Solution
We know that
$x=Acos{\omega}t$
Putting all values
X=6.5 cm
Question 3
Find the maximum
acceleration and maximum velocity
(a) 3.00 cm
2/sec, 6.8 cm/sec
(b) 3.56 cm
2/sec, 6.0 cm/sec
(c) 3.2 cm
2/sec, 6.1 cm/sec
(d) 3.56 cm
2/sec, 6.8 cm/sec
Solution
$v_{max}=\omega A$
=6.8 cm/sec
$a_{max}=\omega^2A$
=3.56 cm2/sec
Question 4
Find the equation of motion of the body
(a) $x=Acos{\omega}t$
(b) $x=Acos{(}\omega t+\pi)$
(c) $x=Acos{(}\omega t-\pi)$
(d) None of these
Solution
At t=0 x=A
So
$x=Acos{\omega}t$
Question 5
Find the force acting on the body when t=2 sec
(a) -64 dyne
(b) -60 dyne
(c) 0 dyne
(d) None of these
Solution
$a=-\omega^2x$
At t=2 x=6.5 cm
Now
F=ma
=-64 dyne
Multiple Choice Questions
Question 6
A solid cylinder is attached to a horizontal massless spring so that it can roll with slipping along the horizontal surface. The spring constant is K .Mass of the cylinder is M.
The system is released from rest where the spring is stretched by x..The
Center of mass of the cylinder execute SHM with time period T.Pick the correct value of T
(a) $T=2\pi\sqrt{\frac{3M}{2K}}$
(b) $T=2\pi\sqrt{\frac{2M}{3K}}$
(c) $T=2\pi\sqrt{\frac{M}{K}}$
(d) None of these
Solution
If at point, the stretch in the spring is x and velocity of center of mass of the cylinder is v
Then
Potential energy=$\frac{1}{2}Kx^2$
Translational kinetic energy
$=\frac{1}{2}Mv^2$
Rotational kinetic energy
$=\frac{1}{2}I\omega^2=\frac{1}{4}Mv^2$
So total energy of the system
$ E=\frac{1}{2}Mv^2+\frac{1}{4}Mv^2+\frac{1}{2}Kx^2$
$E=\frac{3}{4}Mv^2+\frac{1}{2}Kx^2$
Now for SHM
$\frac{dE}{dt}=0$
Solving we get
$T=2\pi\sqrt{\frac{3M}{2K}}$
Question 7
A mass M at the end of a spring executes SHM with a period t
1 while the same mass execute SHM with a period t
2 for another spring. T is the period of oscillation when the two springs are connected in series and Mass M is attached at the end.
Find out the correct relation
(a) $\frac{1}{T}=\frac{1}{t_1}+\frac{1}{t_2}$
(b) $T=t_1+t_2$
(c) $T^2=t_1^2+t_2^2$
(d) $\frac{1}{T^2}=\frac{1}{t_1^2}+\frac{1}{t_2^2}$
Solution
Time period for SHM in spring is given by
$T=2\pi\sqrt{\frac{M}{K}}$
Where K is the spring constant
Let assume $K_1$ and $K_2$ are the spring constant for first and second spring respectively
Then as per given data
$t_1=2\pi\sqrt{\frac{M}{K_1}}$
$t_2=2\pi\sqrt{\frac{M}{K_2}}$
Now for series combination, effective spring constant
$K=\frac{K_1K_2}{K_1+K_2}$
So
$T=2\pi\sqrt{\frac{M(K_1+K_2)}{K_1K_2}}$
$T^2=4\pi^2\frac{M(K_1+K_2)}{K_1K_2}$
So it is clear that
$T^2=t_1^2+t_2^2$
Question 8
Consider a mass –spring system.This system is given an initial displacement ,it begin to oscillate with frequency f
1 .System is now bring to rest and again it is given different displacement and f
2 be its frequency of oscillation then frequencies
(a) f
1 = f
2
(b) f
1 > f
2
(c) f
1 < f
2
(d) none of the above
Solution
(a) f1=f2
Because frequency is not dependent on amplitude of motion
Question 9
The instantaneous displacement of a particle of mass m executing SHM under a force constant k is
$x=Asin{(}\omega t+\varphi)$
Where $\omega=\sqrt{\frac{k}{m}}$
The time average of kinetic energy over a Time period T is
(a) $kA^2$
(b) $\frac{1}{4}kA^2$
(c) $\frac{1}{3}kA^2$
(d) $\frac{1}{2}kA^2$
Solution
Ans is (b)
Kinetic energy
$ K=\frac{1}{2}mv^2=\frac{1}{2}m\omega^2A^2{cos}^2{(}\omega t+\varphi)$
Average KE for one periodic motion is
$K_{avg}=\frac{\int_{0}^{T}Kdt}{\int_{0}^{T}dt}$
$=\frac{1}{T}\int_{0}^{T}{\frac{1}{2}m\omega^2A^2{cos}^2{(}\omega t+\varphi)dt}$
$=\frac{m\omega^2A^2}{2T}\int_{0}^{T}\left(\frac{cos{2}(\omega t+\varphi)+1}{2}\right)dt$
$=\frac{m\omega^2A^2}{2T}\left[\int_{0}^{T}{\frac{1}{2}dt+\int_{0}^{T}{\frac{cos{2}(\omega t+\varphi)}{2}dt}}\right]$
$=\frac{m\omega^2A^2}{4T}\left[T+\left[\frac{sin{2}(\omega t+\varphi)}{2\omega}\right]_0^T\right]$
Now $T=\frac{2\pi}{\omega}$
$=\frac{m\omega^2A^2}{4}$
Now as $\omega=\sqrt{\frac{k}{m}}$
=> $K_{avg}=\frac{1}{4}kA^2$
Question 10
For small amplitude of oscillations potential energy curve w.r.t distance travelled from equilibrium position is
(a) Parabolic
(b) Hyperbolic
(c) Elliptical
(d) circular
Solution
$U=\frac{1}{2}kx^2$
Equation of a parabola is $y^2=4ax$
Here $x^2=\frac {2U}{k}$
so a is correct answer
Question 11
The homogenous linear differential equation
$\frac{d^2x}{dt^2}+2r\frac{dx}{dt}+\omega^2x=0$
Represents the equation of
(a) Simple harmonic oscillator
(b) Damped harmonic oscillator
(c) Forced harmonic oscillator
(d) None of the above
Solution
Answer is (b)
We know equation of damped harmonic oscillator is
$m\ddot{x}+\gamma\dot{x}+kx=0$
=> $\ddot{x}+2\frac{\gamma}{2m}\dot{x}+\frac{k}{m}x=0$
Putting $r=\frac{\gamma}{2m}$
$\omega^2=\frac{k}{m}$
Equation becomes
$\ddot{x}+2r\dot{x}+\omega^2x=0$
Question 12
Given the maximum velocity and acceleration of a harmonic oscillator as v
max and a
max respectively, its time period in terms of v
max and a
max is
(a) $\frac{2\pi v_{max}}{a_{max}}$
(b) $\frac{2\pi a_{max}}{v_{max}}$
(c) $2\pi a_{max}v_{max}$
(d) $\frac{\pi v_{max}}{a_{max}}$
Solution
Answer is (a)
$v_{max}=\omega A$
$a_{max}=\omega^2A$
So $\frac{a_{max}}{v_{max}}=\omega$
Now $\omega=\frac{2\pi}{T}$
So $T=\frac{2\pi v_{max}}{a_{max}}$
Question 13
Which of the following function represents a simple harmonic oscillation
(a) $sin \omega t-cos \omega t$
(b) $sin^2 \omega t$
(c) $sin \omega x+sin 2 \omega t$
(d) $sin \omega x-sin 2 \omega t$
Solution
(a) is the correct answer
Question 14
The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\alpha$, is given by
(a) $T= 2 \pi \sqrt {\frac {L}{g cos \alpha}}$
(b) $T= 2 \pi \sqrt {\frac {L}{g}}$
(c) $T= 2 \pi \sqrt {\frac {L}{g sin \alpha}}$
(d) $T= 2 \pi \sqrt {\frac {L}{g tan \alpha}}$
Solution
(a) is the correct answer
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