Simple Harmonic Motion MCQ

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) $\pm5.28 \ cm/sec$

Answer

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

Answer

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 cm²/sec, 6.8 cm/sec
(b) 3.56 cm²/sec, 6.0 cm/sec
(c) 3.2 cm²/sec, 6.1 cm/sec
(d) 3.56 cm²/sec, 6.8 cm/sec

Answer

$v_{max}=\omega A$
=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

Answer

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

Answer

$a=-\omega^2x$
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

Answer

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 t₁ while the same mass execute SHM with a period t₂ 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) $\frac{1}{T^2}=\frac{1}{t_1^2}+\frac{1}{t_2^2}$

Answer

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 f₁ .System is now bring to rest and again it is given different displacement and f₂ be its frequency of oscillation then frequencies
(a) f₁ = f₂
(b) f₁ > f₂
(c) f₁ < f₂
(d) none of the above

Answer

(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) $\frac{1}{2}kA^2$

Answer

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

Answer

$U=\frac{1}{2}kx^2$

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

Answer

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) $\frac{\pi v_{max}}{a_{max}}$

Answer

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) $sin \omega x-sin 2 \omega t$

Answer

(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) $T= 2 \pi \sqrt {\frac {L}{g tan \alpha}}$

Answer

(a) is the correct answer

---

Also Read

---

• Notes
  • Periodic Motion
  • Simple Harmonic Motion
  • Equation of SHM
  • Characterstics of SHM
  • Velocity of SHM
  • Acceleration of SHM
  • Total Energy of SHM
  • Motion of a body suspended from a spring
  • Simple pendulum
  • The compound pendulum
  • Damped Oscillations
  • Driven or Forced Harmonic oscillator
• Assignments
  • SHM Multiple Choice Questions
  • SHM Problems

---

---

Latest Updates Sound Class 8 Science Quiz

---

Limits Class 11 MCQ

---

Circles in Conic Sections Class 11 MCQ

---

Plant Kingdom free NEET mock tests

---

The Living World free NEET mock tests

---