Simple Harmonic Motion — Page 6 | JEE Physics

Q51

Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is - (consider radius of earth

R_{E}=6400 \mathrm{~km}

and

\mathrm{g}

on earth

10 \mathrm{~m} / \mathrm{s}^{2}

)

A 1200 km

B 1600 km

C 3200 km

D 4800 km

Correct Answer

Option C

Solution

T \propto \sqrt {1/g}

\Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{g_2}} \over {{g_1}}}} = {R \over {R + h}}

{4 \over 6} = {R \over {R + h}}

\Rightarrow h = R/2

= 3200

km

Q52

The time 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

2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \cos \alpha)}

B

2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \sin \alpha)}

C

2 \pi \sqrt{\mathrm{L} / \mathrm{g}}

D

2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \tan \alpha)}

Correct Answer

Option A

Solution

\left| {{g_{eff}}} \right| = \left| {\overline g - \overline a } \right|

\Rightarrow {g_{eff}} = g\cos \theta

\Rightarrow T = 2\pi \sqrt {{l \over {{g_{eff}}}}}

= 2\pi = \sqrt {{L \over {g\cos \theta }}}

Q53

T is the time period of simple pendulum on the earth's surface. Its time period becomes

x

T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of

x

will be :

A 4

B

\dfrac{1}{2}

C 2

D

\dfrac{1}{4}

Correct Answer

Option C

Solution

At surface of earth time period

\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}

At height $\mathrm{h}=\mathrm{R}$

\begin{aligned} & \mathrm{g}^{\prime}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}=\frac{\mathrm{g}}{4} \\\\ & \therefore \,\mathrm{xT}=2 \pi \sqrt{\frac{\ell}{(\mathrm{g} / 4)}} \\\\ & \Rightarrow \mathrm{xT}=2 \times 2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\\\ & \Rightarrow \mathrm{xT}=2 \mathrm{~T} \Rightarrow \mathrm{x}=2 \end{aligned}

Q54

A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is :

A

\dfrac{1}{\sqrt{2}} A

B

\dfrac{1}{2} A

C

2 \mathrm{~A}

D

\sqrt{2 A}

Correct Answer

Option A

Solution

The total energy of a particle executing simple harmonic motion (SHM) is given by:

E = \frac{1}{2}m\omega^2A^2

where

m

is the mass of the particle, $\omega$ is the angular frequency of the SHM, and

A

is the amplitude of the motion. At any point during SHM, the kinetic energy of the particle is given by:

K = \frac{1}{2}m\omega^2(x^2 + A^2\cos^2\omega t)

where

x

is the displacement of the particle from the mean position.

The potential energy of the particle at the same point is given by:

U = \frac{1}{2}m\omega^2(x^2 + A^2\sin^2\omega t)

When the kinetic energy becomes equal to the potential energy, we have:

K = U

\frac{1}{2}m\omega^2(x^2 + A^2\cos^2\omega t) = \frac{1}{2}m\omega^2(x^2 + A^2\sin^2\omega t)

Simplifying this equation, we get:

x^2 = \frac{1}{2}A^2

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

Therefore, the distance from the mean position when the kinetic energy becomes equal to the potential energy is

\boxed{\frac{1}{\sqrt{2}}A}

Q55

A simple pendulum of length

1 \mathrm{~m}

has a wooden bob of mass

1 \mathrm{~kg}

. It is struck by a bullet of mass

10^{-2} \mathrm{~kg}

moving with a speed of

2 \times 10^2 \mathrm{~ms}^{-1}

. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use

\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2

)

A

0.20 \mathrm{~m}

B

0.40 \mathrm{~m}

C

0.30 \mathrm{~m}

D

0.35 \mathrm{~m}

Correct Answer

Option A

Solution

The initial momentum of the system (bullet + bob) is the momentum of the bullet because the bob is initially at rest.

The momentum of the bullet is given by its mass times its velocity :

p_{\text{initial}} = m_{\text{bullet}} \times v_{\text{bullet}}

After the collision, the bullet and the bob move together with a common velocity.

Let's denote this common velocity as $v'$ .

The final momentum $p_{\text{final}}$ is the combined mass of the bullet and bob times the common velocity :

p_{\text{final}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v'

According to the principle of conservation of linear momentum,

p_{\text{initial}} = p_{\text{final}}

m_{\text{bullet}} \times v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v'

Plugging in the values :

(10^{-2} \text{ kg}) \times (2 \times 10^2 \text{ m/s}) = (10^{-2} \text{ kg} + 1 \text{ kg}) \times v'

Solving for $v'$ :

v' = \frac{10^{-2} \times 2 \times 10^2}{10^{-2} + 1} = \frac{2}{1.01} \approx 1.98 \text{ m/s}

After the collision, the system has some kinetic energy which will be completely converted to potential energy at the maximum height $h$ that the bob reaches.

Using the principle of conservation of energy :

\text{Kinetic Energy (KE)}_{\text{initial}} = \text{Potential Energy (PE)}_{\text{final}}

\frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2 = (m_{\text{bullet}} + m_{\text{bob}})gh

Isolating $h$ , we get :

h = \frac{\frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2}{(m_{\text{bullet}} + m_{\text{bob}})g} = \frac{v'^2}{2g}

Substituting the values for $v'$ and $g$ :

h = \frac{(1.98)^2}{2 \times 10} = \frac{3.9204}{20} = 0.19602 \text{ m}

Looking at the given options, the result most closely matches Option A :

\boxed{0.20 \text{ m}}

Therefore, the height to which the bob rises before swinging back is approximately $0.20 \text{ m}$ .

Q56

A simple pendulum doing small oscillations at a place

R

height above earth surface has time period of

T_1=4 \mathrm{~s}

.

\mathrm{T}_2

would be it's time period if it is brought to a point which is at a height

2 \mathrm{R}

from earth surface. Choose the correct relation [

\mathrm{R}=

radius of earth] :

A

3 \mathrm{~T}_1=2 \mathrm{~T}_2

B

\mathrm{T}_1=\mathrm{T}_2

C

2 \mathrm{~T}_1=3 \mathrm{~T}_2

D

2 \mathrm{~T}_1=\mathrm{T}_2

Correct Answer

Option A

Solution

The time period of a simple pendulum is given by the formula:

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

where

T

is the time period,

l

is the length of the pendulum, and

g

is the acceleration due to gravity at the location of the pendulum.

The acceleration due to gravity changes with height above the Earth's surface.

The acceleration due to gravity at a height

h

above the Earth's surface can be expressed as:

g' = g \left(\frac{R}{R + h}\right)^2

where

g

is the acceleration due to gravity at the surface of the Earth,

R

is the radius of the Earth, and

h

is the height above the Earth’s surface.

Since the time period of the pendulum depends on the square root of the inverse of the acceleration due to gravity, any change in

g

due to a change in height will affect the time period. Given that the time period of the pendulum at a height

R

above Earth's surface is

T_1

, and we're to find the time period

T_2

at a height of

2R

, we can use the formula for acceleration due to gravity at different heights to express the relationship between

T_1

and

T_2

. For the initial case at height

R

:

g_1 = g \left(\frac{R}{R + R}\right)^2 = g \left(\frac{R}{2R}\right)^2 = \frac{g}{4}

For the new case at height

2R

:

g_2 = g \left(\frac{R}{R + 2R}\right)^2 = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}

The time period is proportional to the square root of the inverse of

g

, so:

\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac{\frac{g}{9}}{\frac{g}{4}}} = \sqrt{\frac{4}{9}} = \frac{2}{3}

Therefore:

T_1 = \frac{2}{3}T_2

Rearranging this equation:

3T_1 = 2T_2

This corresponds to Option A.

Q57

A particle is subjected to two simple harmonic motions as :

x_1=\sqrt{7} \sin 5 \mathrm{tcm}

and

x_2=2 \sqrt{7} \sin \left(5 t+\dfrac{\pi}{3}\right) \mathrm{cm}

where

x

is displacement and

t

is time in seconds. The maximum acceleration of the particle is

x \times 10^{-2} \mathrm{~ms}^{-2}

. The value of

x

is :

A

5 \sqrt{7}

B 125

C

25 \sqrt{7}

D 175

Correct Answer

Option D

Solution

\begin{aligned} &\begin{aligned} & x_1=\sqrt{7} \sin 5 t \\ & x_2=2 \sqrt{7} \sin \left(5 t+\frac{\pi}{3}\right) \end{aligned}\\ &\text{ From phasor, } \end{aligned}

\begin{aligned} &\therefore \text{ Amplitude of resultant } \mathrm{SHM}=7\\ &\begin{aligned} & \phi=\tan ^{-1} \frac{2 \sqrt{7} \times \sqrt{3} / 2}{\sqrt{7}+2 \sqrt{7} \times \frac{1}{2}}=\tan ^{-1} \frac{\sqrt{21}}{2 \sqrt{7}}=\tan ^{-1} \frac{\sqrt{3}}{2} \\ & \therefore X_R=7 \sin (5 \mathrm{t}+\phi) \\ & \quad a_R=-7 \times 25 \sin (5 \mathrm{t}+\phi) \\ & \therefore a_{\max }=175 \mathrm{~cm} / \mathrm{sec}=175 \times 10^{-2} \mathrm{~m} / \mathrm{sec} \end{aligned} \end{aligned}

Q58

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. Reason (R) : The mass of the pendulum remains unchanged at Earth and the other planet. In the light of the above statements, choose the correct answer from the options given below :

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B (A) is false but (R) is true

C (A) is true but (R) is false

D Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

Correct Answer

Option D

Solution

Assertion (A): A simple pendulum transported to a planet where the mass and radius are 4 times and 2 times that of the Earth, respectively, has the same time period as it does on Earth.

Reason (R): The mass of the pendulum remains unchanged whether on Earth or the other planet.

Explanation: The acceleration due to gravity $g$ on a planet is given by the formula: $g = \dfrac{G M}{R^2}$ where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.

For the new planet, the gravitational acceleration $g'$ is: $g' = \dfrac{G(4M)}{(2R)^2} = \dfrac{4G M}{4R^2} = \dfrac{G M}{R^2} = g$ Thus, the gravitational acceleration on this new planet is the same as on Earth, $g$ .

The time period $T$ of a simple pendulum is determined by: $T = 2\pi \sqrt{\dfrac{\ell}{g}}$ where $\ell$ is the length of the pendulum, which indicates that the time period $T$ is independent of the mass of the pendulum.

Therefore, while Assertion (A) is true and Reason (R) is true, Reason (R) does not correctly explain Assertion (A) because the time period of the pendulum is also independent of the pendulum's mass, and the key factor is the unchanged gravitational acceleration.

Q59

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Knowing initial position

\mathrm{x}_0

and initial momentum

p_0

is enough to determine the position and momentum at any time

t

for a simple harmonic motion with a given angular frequency

\omega

. Reason (R) : The amplitude and phase can be expressed in terms of

\mathrm{X}_0

an

\mathrm{p}_0

. In the light of the above statements, choose the correct answer from the options given below :

A (A) is true but (R) is false

B Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

C (A) is false but (R) is true

D Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\phi) \\ & \mathrm{x}_0=\mathrm{A} \sin \phi \quad\text{.... (1)}\\ & \mathrm{p}=\mathrm{mA} \omega \cos (\omega \mathrm{t}+\phi) \\ & \mathrm{p}_0=\mathrm{mA} \omega \cos \phi \quad\text{.... (2)} \end{aligned}

$(2) /(1) \Rightarrow \tan \phi=\left(\dfrac{x_0}{p_0}\right) m \omega$

\sin \phi=\frac{\mathrm{x}_0 \mathrm{~m} \omega}{\sqrt{\left(\mathrm{~m} \omega \mathrm{x}_0\right)^2+\mathrm{p}_0^2}}

From (1), $A=\dfrac{x_0}{\sin \phi}=\dfrac{\sqrt{\left(m \omega x_0\right)^2+p_0^2}}{m \omega}$ This means we can explain assertion with the given reason.

Q60

A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is

y \pi \times 10^{-2} \mathrm{~s}

, where the value of

y

is (Acceleration due to gravity,

g=10 \mathrm{~m} / \mathrm{s}^2

, density of water

=10^3 \mathrm{~kg} / \mathrm{m}^3

)

A 2

B 4

C 1

D 6

Correct Answer

Option A

Solution

We can determine the time period of the oscillations by considering that when the cube is depressed by a small displacement, the additional buoyant force provided by the displaced water acts as a restoring force.

Here’s a step‐by‐step explanation: Define the given values: Side length of the cube,

L = 10\text{ cm} = 0.1\text{ m}

. Mass of the cube,

m = 10\text{ g} = 0.01\text{ kg}

. Density of water,

\rho = 10^3\ \text{kg/m}^3

. Acceleration due to gravity,

g = 10\ \text{m/s}^2

. Cross-sectional area of the cube (face area),

A = L^2 = (0.1)^2 = 0.01\text{ m}^2.

When the cube is depressed by a small distance

x

, the additional volume of water displaced is

\Delta V = A\,x.

Thus, the additional buoyant force is:

F_b = \rho\,g\,\Delta V = \rho\,g\,A\,x.

This force acts in the upward (restoring) direction. Notice that the force is proportional to the displacement

x

, which is the hallmark of simple harmonic motion (SHM). The effective spring constant,

k

, associated with this SHM is:

k = \rho\,g\,A.

The period of oscillation for SHM is given by:

T = 2\pi \sqrt{\frac{m}{k}},

which, upon substituting for

k

, becomes:

T = 2\pi \sqrt{\frac{m}{\rho\,g\,A}}.

Substitute the given values into the expression:

T = 2\pi \sqrt{\frac{0.01}{1000 \times 10 \times 0.01}}.

Calculate the denominator:

1000 \times 10 \times 0.01 = 100,

so,

T = 2\pi \sqrt{\frac{0.01}{100}} = 2\pi \sqrt{0.0001}.

Since,

\sqrt{0.0001} = 0.01,

we have:

T = 2\pi \times 0.01 = 0.02\pi\ \text{s}.

The problem states that the time period is given by:

y \pi \times 10^{-2}\ \text{s}.

Writing our result in the same form:

0.02\pi\ \text{s} = 2\pi \times 10^{-2}\ \text{s}.

We see that

y = 2.

Therefore, the correct answer is: Option A: 2.