Simple Harmonic Motion Questions — NEET Physics

Q1

The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately: (Consider mass of the bob

=20 \mathrm{~g}

)

A

0.2 \mathrm{~m} / \mathrm{s}

B

1.41 \mathrm{~m} / \mathrm{s}

C

14.1 \mathrm{~m} / \mathrm{s}

D

2.0 \mathrm{~m} / \mathrm{s}

Correct Answer

Option B

Solution

At equilibrium position Total energy = K.E

\begin{aligned} & \frac{1}{2} m v^2=0.02 \\ & \frac{1}{2} \times 20 \times v^2 \times 10^{-3}=2 \times 10^{-2} \\ & v=\sqrt{2} \\ & v=1.41 \mathrm{~m} / \mathrm{s} \end{aligned}

Q2

Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum

L

, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (Take

\pi^2=9.8

, and

g=9.8 \mathrm{~m} / \mathrm{s}^2

)

A 0.75 m

B 1.5 m

C 2 m

D 1 m

Correct Answer

Option D

Solution

Time taken for 30 oscillations $=60 \mathrm{~s}$ Time period of simple pendulum = Time taken for 1 oscillation

\begin{aligned} & \Rightarrow T=\frac{60}{30}=2 \mathrm{~s} \\ & T=2 \pi \sqrt{\frac{I}{g}} \Rightarrow I=\frac{g T^2}{4 \pi^2}=\frac{9.8 \times 2 \times 2}{4 \times 9.8}=1 \mathrm{~m} \end{aligned}

Q3

Two identical point masses

P

and

Q

, suspended from two separate massless springs of spring constants

k_1

and

k_2

, respectively, oscillate vertically. If their maximum speeds are the same, the ratio (

A Q / A_\rho

) of the amplitude

A Q

of mass

Q

to the amplitude

A_\rho

of mass

P

is

A

\sqrt{\dfrac{k_2}{k_1}}

B

\sqrt{\dfrac{k_1}{k_2}}

C

\dfrac{k_2}{k_1}

D

\dfrac{k_1}{k_2}

Correct Answer

Option B

Solution

Two identical point masses, $P$ and $Q$ , are suspended from two separate massless springs with spring constants $k_1$ and $k_2$ , respectively.

These masses oscillate vertically, and it is given that their maximum speeds are the same.

We need to determine the ratio of the amplitude $A_Q$ of mass $Q$ to the amplitude $A_P$ of mass $P$ .

The maximum velocity $V$ for an oscillating mass is given by the equation $V = A \omega$ , where $A$ is the amplitude and $\omega$ is the angular frequency.

Given that the maximum velocities of $P$ and $Q$ are the same: $v_P = v_Q$ This implies: $A_P \omega_P = A_Q \omega_Q$ From this, the ratio of the amplitudes can be expressed as: $\dfrac{A_Q}{A_P} = \dfrac{\omega_P}{\omega_Q}$ The angular frequency $\omega$ of a mass-spring system is given by: $\omega = \sqrt{\dfrac{k}{m}}$ Thus, for the two masses: $\omega_P = \sqrt{\dfrac{k_1}{m}} \quad \text{and} \quad \omega_Q = \sqrt{\dfrac{k_2}{m}}$ Substituting these into the ratio equation, we get: $\dfrac{A_Q}{A_P} = \dfrac{\sqrt{\dfrac{k_1}{m}}}{\sqrt{\dfrac{k_2}{m}}} = \sqrt{\dfrac{k_1}{k_2}}$ Hence, the ratio of the amplitude of mass $Q$ to the amplitude of mass $P$ is: $\sqrt{\dfrac{k_1}{k_2}}$

Q4

A particle executing simple harmonic motion with amplitude A has the same potential and kinetic energies at the displacement

A

2 \sqrt{A}

B

\dfrac{A}{2}

C

\dfrac{A}{\sqrt{2}}

D

A \sqrt{2}

Correct Answer

Option C

Solution

In simple harmonic motion (SHM), the total mechanical energy of the system is conserved and is a combination of kinetic energy (KE) and potential energy (PE).

The total energy (E) of the particle can be expressed as:

E = \frac{1}{2}kA^2,

where $k$ is the spring constant, and $A$ is the amplitude of motion.

At a displacement $x$ , the potential energy (PE) and kinetic energy (KE) of the particle are given by:

\text{PE} = \frac{1}{2}kx^2

\text{KE} = E - \text{PE} = \frac{1}{2}kA^2 - \frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2)

We need to find the displacement where the potential and kinetic energies are equal. This implies:

\text{PE} = \text{KE}

Thus, we get:

\frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2)

By simplifying, we get:

x^2 = A^2 - x^2

Adding $x^2$ to both sides:

2x^2 = A^2

Now, solving for $x$ :

x = \frac{A}{\sqrt{2}}

Therefore, the displacement at which the potential and kinetic energies are equal is: Option C:

\frac{A}{\sqrt{2}}

Q5

If

x=5 \sin \left(\pi t+\dfrac{\pi}{3}\right) \mathrm{m}

represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

A 5 cm, 2 s

B 5 m, 2 s

C 5 cm, 1 s

D 5 m, 1 s

Correct Answer

Option B

Solution

In the equation for simple harmonic motion (SHM),

x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m},

the general form

x = A \sin (\omega t + \phi)

can be used to identify the parameters of SHM, where: A is the amplitude. ω (omega) is the angular frequency. φ (phi) is the phase constant. t is the time.

Comparing the given equation with the standard form: The amplitude A is 5 m, as that is the coefficient of sine in the equation.

The angular frequency ω is $\pi$ rad/s.

The angular frequency $\omega$ is related to the time period

T

of the motion through the formula:

\omega = \frac{2\pi}{T}

. Given that

\omega = \pi

, we can substitute and solve for

T

:

\begin{aligned} \pi = \frac{2\pi}{T} & \implies T = \frac{2\pi}{\pi} = 2 \text{ seconds}. \end{aligned}

Hence, the amplitude of the motion is 5 m and the time period is 2 s. Thus, the correct answer is: Option B: 5 m, 2 s

Q6

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is

\dfrac{x}{2}

times its original time period. Then the value of

x

is:

A

\sqrt{3}

B

\sqrt{2}

C

2 \sqrt{3}

D 4

Correct Answer

Option B

Solution

The period of oscillation,

T

, of a simple pendulum is determined by the formula:

T = 2\pi \sqrt{\frac{L}{g}}

where:

L

is the length of the pendulum

g

is the acceleration due to gravity The mass of the bob does not factor into the equation for the period.

Let's first denote the original length of the pendulum as

L

and the original period of oscillation as

T_1

. Hence,

T_1 = 2\pi \sqrt{\frac{L}{g}}

When the length of the pendulum is halved, the new length

L'

would be

\frac{L}{2}

. Thus, the new period

T_2

can be calculated as:

T_2 = 2\pi \sqrt{\frac{\frac{L}{2}}{g}} = 2\pi \sqrt{\frac{L}{2g}} = 2\pi \left(\frac{1}{\sqrt{2}}\right) \sqrt{\frac{L}{g}} = \frac{1}{\sqrt{2}} \cdot 2\pi \sqrt{\frac{L}{g}} = \frac{T_1}{\sqrt{2}}

We are given that the new period

T_2

is

\frac{x}{2} T_1

. Therefore, we can set up the equation:

\frac{T_1}{\sqrt{2}} = \frac{x}{2} T_1

To find the value of

x

, we solve for

x

:

\frac{1}{\sqrt{2}} = \frac{x}{2}

Multiplying both sides by 2:

\frac{2}{\sqrt{2}} = x

Simplify to:

x = \sqrt{2}

Hence, the correct answer is: Option B:

\sqrt{2}

Q7

A simple pendulum oscillating in air has a period of

\sqrt{3} \mathrm{~s}

. If it is completely immersed in non-viscous liquid, having density

\left(\dfrac{1}{4}\right)^{\text{th }}

of the material of the bob, the new period will be :-

A

2 \sqrt{3}

s

B

\dfrac{2}{\sqrt{3}} \mathrm{~s}

C

2 \mathrm{~s}

D

\dfrac{\sqrt{3}}{2} \mathrm{~s}

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{T}_{\text{air }}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}=\sqrt{3} ~\mathrm{sec} \\ & \ln \text{ Liquid } \rightarrow g_{\text{net }}=g\left(1-\frac{\rho}{\sigma}\right) \\ & \rho=\text{ density of liquid } \\ & \sigma=\text{ density of material of bob } \\ & \text{ so } \mathrm{T}_{\text{Liq }}=2 \pi \sqrt{\left(\frac{\ell}{g_{\text{net }}}\right)}=2 \pi \sqrt{\frac{\ell}{g\left(1-\frac{\rho}{\sigma}\right)}} \\ & \mathrm{T}_{\text{Liq }}=\frac{\mathrm{T}_{\text{air }}}{\sqrt{1-\frac{\rho}{\sigma}}}=\frac{\sqrt{3}}{\sqrt{1-\frac{1}{4}}}=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=2 ~\mathrm{sec} \end{aligned}

Q8

The

x

-

t

graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at

t=2 \mathrm{~s}

is :

A

-\dfrac{\pi^{2}}{8} \mathrm{~ms}^{-2}

B

\dfrac{\pi^{2}}{16} \mathrm{~ms}^{-2}

C

-\dfrac{\pi^{2}}{16} \mathrm{~ms}^{-2}

D

\dfrac{\pi^{2}}{8} \mathrm{~ms}^{-2}

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}) \\ & \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}=\mathrm{A} \omega \cos (\omega \mathrm{t}) \\ & \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{a}=-\omega^{2} \mathrm{~A} \sin (\omega \mathrm{t}) \\ & \mathrm{a}=-\left(\frac{2 \pi}{8}\right)^{2} \times 1 \sin \left(\frac{2 \pi}{8} \times 2\right) \\ & \Rightarrow \mathrm{a}=-\frac{\pi^{2}}{16} \times \sin \left(\frac{\pi}{2}\right) \\ & \therefore \mathrm{a}=\frac{-\pi^{2}}{16} \mathrm{~m} / \mathrm{s}^{2} \end{aligned}

Q9

Identify the function which represents a non-periodic motion.

A

\sin (\omega t + \pi /4)

B

{e^{ - \omega t}}

C

\sin \omega t

D

\sin \omega t + \cos \omega t

Correct Answer

Option B

Solution

Here,

\sin \omega t

and

\sin \left( {\omega t + {\pi \over 4}} \right)

clearly suggest a periodic sinusoidal function (SHM) And

\sin \omega t + \cos \omega t = \sqrt 2 \sin \left( {\omega t + {\pi \over 4}} \right)

is also a periodic sinusoidal function (SHM) While,

y = {e^{ - \omega t}}

is an exponentially decreasing function and thus does not have periodicity.

Q10

Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :

A 11

B 9

C 10

D 8

Correct Answer

Option A

Solution

T = 2\pi \sqrt {{L \over g}}

Let n 1 and n 2 be integer.

{n_1}{T_1} = {n_2}{T_2}

2\pi {n_1}\sqrt {{{1.21} \over g}} = 2\pi {n_2}\sqrt {{{1.00} \over g}}

\Rightarrow {{{n_2}} \over {{n_1}}} = {{11} \over {10}}

$\therefore$ After completion of 11 th oscillation of shorter pendulum, it will be in phase with longer pendulum.