Simple Harmonic Motion — Page 11 | JEE Physics

Q101

A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm−1 and oscillates in a damping medium of damping constant 10−2 kg s−1 . The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to :

A 2 s

B 3.5 s

C 5 s

D 7 s

Correct Answer

Option D

Solution

To determine the time taken for the mechanical energy of the damped oscillator to drop to half its initial value, we'll use the principles of damped harmonic motion.

Given: Mass of the block ( $m$ ): 0.1 kg Spring constant ( $k$ ): 640 N/m Damping constant ( $b$ ): $10^{-2}$ kg/s Understanding Damped Harmonic Motion: In damped harmonic motion, the amplitude of oscillation decreases exponentially over time due to the damping force.

The mechanical energy ( $E$ ) of the oscillator is proportional to the square of its amplitude ( $A$ ): $E(t) \propto A(t)^2$ The amplitude as a function of time is given by: $A(t) = A_0 \, e^{- \dfrac{b}{2m} t}$ Where: $A_0$ is the initial amplitude. $b$ is the damping constant. $m$ is the mass. $t$ is the time.

Therefore, the mechanical energy as a function of time is: $E(t) = E_0 \, e^{- \dfrac{b}{m} t}$ Where $E_0$ is the initial mechanical energy.

Calculating the Time When Energy Drops to Half: We need to find the time $t$ when $E(t) = \dfrac{1}{2} E_0$ : $\dfrac{1}{2} E_0 = E_0 \, e^{- \dfrac{b}{m} t}$ Simplify: $\dfrac{1}{2} = e^{- \dfrac{b}{m} t}$ Take the natural logarithm of both sides: $\ln\left(\dfrac{1}{2}\right) = - \dfrac{b}{m} t$ Simplify $\ln\left(\dfrac{1}{2}\right) = -\ln(2)$ : $- \ln(2) = - \dfrac{b}{m} t$ Cancel negatives: $\ln(2) = \dfrac{b}{m} t$ Solve for $t$ : $t = \dfrac{m}{b} \ln(2)$ Plugging in the Given Values: $t = \dfrac{0.1\, \text{kg}}{10^{-2}\, \text{kg/s}} \ln(2)$ Calculate $\ln(2)$ : $\ln(2) \approx 0.6931$ Now compute $t$ : $t = \dfrac{0.1}{0.01} \times 0.6931 = 10 \times 0.6931 = 6.931\, \text{s}$ Conclusion: The time taken for the mechanical energy to drop to half its initial value is approximately 6.93 seconds, which is closest to 7 seconds among the given options.

Q102

The maximum potential energy of a block executing simple harmonic motion is

25 \mathrm{~J}

. A is amplitude of oscillation. At

\mathrm{A / 2}

, the kinetic energy of the block is

A 9.75 J

B 37.5 J

C 18.75 J

D 12.5 J

Correct Answer

Option C

Solution

$\mathrm{U}_{\max }=\dfrac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}=25 \mathrm{~J}$ $\mathrm{KE}$ at $\dfrac{\mathrm{A}}{2}=\dfrac{1}{2} m v_{1}^{2}=\dfrac{1}{2} m \omega^{2}\left(A^{2}-\dfrac{A^{2}}{4}\right)$ $=\dfrac{1}{2} \mathrm{~m} \omega^{2} \dfrac{3 \mathrm{~A}^{2}}{4}=\dfrac{3}{4}\left(\dfrac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\right)$ $\mathrm{KE}=\dfrac{3}{4} \times 25=18.75 \mathrm{~J}$

Q103

The displacement of an object attached to a spring and executing simple harmonic motion is given by

x = 2 \times {10^{ - 2}}

cos

\pi t

metre. The time at which the maximum speed first occurs is

A

0.25

s

B

0.5

s

C

0.75

s

D

0.125

s

Correct Answer

Option B

Solution

Here,

x = 2 \times {10^{ - 2}}\cos \,\pi \,t

$\therefore$

v = {{dx} \over {dt}} = 2 \times {10^{ - 2}}\,\pi \sin \pi t

For the first time, the speed to be maximum,

\sin \pi t = 1

or,

\sin \pi t = \sin {\pi \over 2}

\Rightarrow \pi t = {\pi \over 2}\,\,\,

or,

\,\,\,\,t = {1 \over 2} = 0.5\,\sec .

Q104

A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

A 1 : 1

B 1 : 4

C 2 : 1

D 1 : 3

Correct Answer

Option D

Solution

Let's denote the amplitude of the simple harmonic motion as A, and the displacement of the particle from the mean position as x.

The given condition is that x = A/2.

For a particle in SHM, the potential energy (PE) is given by:

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

And the total mechanical energy (E) of the particle remains constant and is given by:

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

Since the total mechanical energy is the sum of potential energy and kinetic energy (KE), we have:

E = PE + KE

Now, we need to find the ratio of potential energy to kinetic energy when x = A/2.

Calculate the potential energy at x = A/2:

PE = \frac{1}{2} k\left(\frac{A}{2}\right)^2 = \frac{1}{8} kA^2

Substitute the expression for total mechanical energy:

KE = E - PE = \frac{1}{2} kA^2 - \frac{1}{8} kA^2 = \frac{3}{8} kA^2

Now, find the ratio of potential energy to kinetic energy:

\frac{PE}{KE} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}

Therefore, the ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude is 1 : 3.

Q105

Motion of a particle in x-y plane is described by a set of following equations

x = 4\sin \left( {{\pi \over 2} - \omega t} \right)\,m

and

y = 4\sin (\omega t)\,m

. The path of the particle will be :

A circular

B helical

C parabolic

D elliptical

Correct Answer

Option A

Solution

x = 4\sin \left( {{\pi \over 2} - \omega t} \right)

= 4\cos (\omega t)

y = 4\sin (\omega t)

\Rightarrow {x^2} + {y^2} = {4^2}

$\Rightarrow$ The particle is moving in a circular motion with radius of 4 m.

Q106

A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1/1000 of the original amplitude is close to :-

A 100 s

B 10 s

C 20 s

D 50 s

Correct Answer

Option C

Solution

Time for 10 oscillations =

{{10} \over 5} = 2\,s

A = A0 e–kt

{1 \over 2} = {e^{ - 2k}} \Rightarrow \ln 2 = 2k

10–3 = e–kt $\Rightarrow$ 3In10 = kt

t = {{3\ln 10} \over k} = {{3\ln 10} \over {\ln 2}} \times 2

=

6 \times {{2.3} \over {0.69}} \approx 20\,s

Q107

Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be :

A

\sqrt 3 T

B

\sqrt {{2 \over 3}} T

C

{T \over {\sqrt 3 }}

D

\sqrt {{3 \over 2}} T

Correct Answer

Option B

Solution

When lift is stationary

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

A pseudo force will act downwards when lift is moving upwards. $\therefore$

{g_{eff}} = g + {g \over 2} = {{3g} \over 2}

$\therefore$ New time period

T' = 2\pi \sqrt {{L \over {{g_{eff}}}}}

T' = 2\pi \sqrt {{{2L} \over {3g}}}

$\therefore$

T' = \sqrt {{2 \over 3}} T

Q108

If a simple harmonic motion is represented by

{{{d^2}x} \over {d{t^2}}} + \alpha x = 0.

its time period is

A

{{2\pi } \over {\sqrt \alpha }}

B

{{2\pi } \over \alpha }

C

2\pi \sqrt \alpha

D

2\pi \alpha

Correct Answer

Option A

Solution

{{{d^2}x} \over {d{t^2}}} = - \alpha x = - {\omega ^2}x

\Rightarrow \omega = \sqrt \alpha

\,\,\,\,

or

\,\,\,\,

T = {{2\pi } \over \omega } = {{2\pi } \over {\sqrt \alpha }}