Simple Harmonic Motion — Page 9 | NEET Physics

Q81

A particle moves with simple harmonic motion in a straight line. In first

\tau s,

after starting from rest it travels a distance

a,

and in next

\tau s

it travels

2a,

in same direction, then:

A amplitude of motion is

3a

B time period of oscillations is

8\tau

C amplitude of motion is

4a

D time period of oscillations is

6\tau

Correct Answer

Option D

Solution

In simple harmonic motion, starting from rest, At

t=0,

x=A

x = A\cos \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

When

t = \tau ,\,\,x = A - a

When

t = 2\,\tau ,\,x = A - 3a

From equation

(i)

A - a = A\cos \omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

A - 3a = A\cos 2\omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)

As

\cos 2\omega \,\tau = 2{\cos ^2}\omega \tau - 1...\left( {iv} \right)

From equation

(ii),

(iii)

and

(iv)

{{A - 3A} \over A} = 2{\left( {{{A - a} \over A}} \right)^2} - 1

\Rightarrow {{A - 3a} \over A} = {{2{A^2} + 2{a^2} - 4Aa - {A^2}} \over {{A^2}}}

\Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa

\Rightarrow 2{a^2} = aA \Rightarrow \,\,\,\,\,\,\,A = 2a

\Rightarrow {a \over A} = {1 \over 2}

Now,

A-a=A

\cos \omega \tau

\Rightarrow \cos \omega \tau = {{A - a} \over A} \Rightarrow \,\,\cos \omega \tau = {1 \over 2}

or,

{{2\pi } \over T}\tau = {\pi \over 3} \Rightarrow \,\,\,T - 6\,\tau

Q82

In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

A

{1 \over 2}

B

{3 \over 4}

C

{1 \over 3}

D

{1 \over 4}

Correct Answer

Option B

Solution

K = {1 \over 2}m{\omega ^2}({A^2} - {x^2})

= {1 \over 2}m{\omega ^2}\left( {{A^2} - {{{A^2}} \over 4}} \right)

= {1 \over 2}m{\omega ^2}\left( {{{3{A^2}} \over 4}} \right)

K = {3 \over 4}\left( {{1 \over 2}m{\omega ^2}{A^2}} \right)

Q83

The bob of a simple pendulum executes simple harmonic motion in water with a period

t,

while the period of oscillation of the bob is

{t_0}

in air. Neglecting frictional force of water and given that the density of the bob is

\left( {4/3} \right) \times 1000\,\,kg/{m^3}.

What relationship between

t

and

{t_0}

is true

A

t = 2{t_0}

B

t = {t_0}/2

C

t = {t_0}

D

t = 4{t_0}

Correct Answer

Option A

Solution

t = 2\pi \sqrt {{\ell \over {{g_{eff}}}}} ;\,{t_o}\,\, = 2\pi \sqrt {{\ell \over g}}

m{g_{eff}} = mg - B = my - V \times 100 \times g

$\therefore$

{g_{eff}} = g - {{100} \over {\left( {m/v} \right)}}g

= g - {{1000} \over {{4 \over 3} \times 1000}}g = {g \over 4}

$\therefore$

t = 2\pi \sqrt {{\ell \over {g/4}}} \,\,\,\,\,\,\,\,\,\,\,t = 2{t_0}

Q84

For a body executing S.H.M. : (1) Potential energy is always equal to its K.E. (2) Average potential and kinetic energy over any given time interval are always equal. (3) Sum of the kinetic and potential energy at any point of time is constant. (4) Average K.E. in one time period is equal to average potential energy in one time period. Choose the most appropriate option from the options given below :

A (3) and (4)

B only (3)

C (2) and (3)

D only (2)

Correct Answer

Option A

Solution

In S.H.M. total mechanical energy remains constant and also = =

{{1 \over 4}}

KA2 (for 1 time period)

Q85

An oscillator of mass M is at rest in its equilibrium position in a potential V =

{1 \over 2}

k(x

-

X)2. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is : (M = 10, m = 5, u = 1, k = 1)

A

{1 \over {\sqrt 3 }}

B

{1 \over 2}

C

{2 \over 3}

D

{3 \over {\sqrt 5 }}

Correct Answer

Option A

Solution

Potential of the given oscillator is

V = {1 \over 2}k{(x - k)^2}

Given: M = 10; m = 5, u = 1; k = 1 Initial momentum of the particle of mass m = mu = m $\times$ 5 = 5m Momentum of (oscillator + particle) after collision = (M + m) Velocity of oscillator after collision = v So, momentum of system = (M + m)v From conservation of linear momentum, we have (M + m) = mu = 5 $\times$ 1 = 5 For second collision, oscillator and particle have momentum in opposite direction.

Net or total momentum is zero.

Likewise after 4th, 6th, 8th, 10th, 12th collision the momentum is zero.

After 12th collision, Mass of oscillator and 12 particles will be (10 + 12 $\times$ 5) = 70 Now, from conservation of linear momentum, for 13th collision, we have

70 \times 0 + 5 \times 1 = (70 + 5)v' \Rightarrow v' = {5 \over {75}} \Rightarrow {1 \over {15}}

Total mass after 13th collision = (10 + 13 $\times$ 5) = 75 Kinetic energy of system

= {1 \over 2}mv{'^2}

\Rightarrow KE = {1 \over 2} \times 75 \times {1 \over {15}} \times {1 \over {15}}

\Rightarrow {1 \over 2}k{A^2} = {1 \over 2} \times {{75} \over {225}} = {1 \over 6}

\Rightarrow {1 \over 2} \times 1 \times {A^2} = {1 \over 6} \Rightarrow {A^2} = {1 \over 3}

\Rightarrow A = {1 \over {\sqrt 3 }}

Q86

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz)

A 4

B 7

C 6

D 5

Correct Answer

Option B

Solution

For closed organ pipe, resonate frequency is odd multiple of fundamental frequency. $\therefore$ (2n + 1) f0 $\le$ 20,000 (f0 is fundamental frequency = 1.5 KHz) $\therefore$ n = 6 $\therefore$ Total number of overtone that can be heared is 7. (0 to 6)

Q87

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.

Q88

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

Q89

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 .

Q90

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.