Simple Harmonic Motion — Page 7 | JEE Physics

Q61

A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm . If D and d are the total distance and displacement covered by the particle in 12.5 s , then

\dfrac{\mathrm{D}}{\mathrm{d}}

is

A

10

B

\dfrac{16}{5}

C

25

D

\dfrac{15}{4}

Correct Answer

Option C

Solution

x(t) = \cos(\pi t)

The particle executes simple harmonic motion (SHM) with amplitude

A = 1 \text{ cm}

and period

T = 2 \text{ s}

. In one complete cycle (2 s), the motion is as follows: It starts at

x = 1 \text{ cm}

. It moves to

x = -1 \text{ cm}

(covering a distance of

2 \text{ cm}

). It returns to

x = 1 \text{ cm}

(covering another

2 \text{ cm}

). Thus, the total distance traveled in one complete cycle is:

D_{\text{cycle}} = 2 + 2 = 4 \text{ cm}.

In 12.5 s, the number of cycles is:

\frac{12.5}{2} = 6.25 \text{ cycles}.

For the 6 complete cycles (12 s), the distance traveled is:

D_{\text{complete}} = 6 \times 4 = 24 \text{ cm}.

For the remaining 0.5 s, determine the displacement. At

t = 12 \text{ s}

, the particle is at:

x(12) = \cos(12\pi) = 1 \text{ cm}.

After an extra 0.5 s (i.e., at

t = 12.5 \text{ s}

), the position becomes:

x(12.5) = \cos(12.5\pi) = \cos\left(12\pi + \frac{\pi}{2}\right) = \cos\frac{\pi}{2} = 0 \text{ cm}.

The distance covered in this interval is:

D_{\text{extra}} = |1 - 0| = 1 \text{ cm}.

So, the total distance traveled is:

D = D_{\text{complete}} + D_{\text{extra}} = 24 + 1 = 25 \text{ cm}.

The net displacement, which is the difference between the final and initial positions, is calculated as follows: Initial position at

t = 0

:

x(0) = \cos 0 = 1 \text{ cm},

Final position at

t = 12.5 \text{ s}

:

x(12.5) = 0 \text{ cm}.

Thus, the displacement is:

d = |0 - 1| = 1 \text{ cm}.

The ratio of the total distance traveled to the displacement is:

\frac{D}{d} = \frac{25}{1} = 25.

Therefore, the correct answer is

25

.

Q62

A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance x m (x < 2) from the wall, the speed of the block will be

A

10\left[1-(2-x)^2\right]^{\dfrac{1}{2}} \ m/s

B

10\left[1-(2-x)^2\right]^{\dfrac{3}{2}} \ m/s

C

10\left[1-(2-x)^2\right] \ m/s

D

10\left[1-(2-x)^2\right]^2 \ m/s

Correct Answer

Option A

Solution

Given: Natural length of the spring: 2 meters Initial compression of the spring ( $x_i$ ): 1 meter Final compression of the spring when the block is at distance $x$ ( $x_f$ ): $(2 - x)$ meters Solution: To find the block's speed, we'll use energy conservation.

Initially, the spring's potential energy is fully converted to kinetic energy and spring potential energy at the new position $x$ .

Conservation of Energy: Initial kinetic energy ( $K_i$ ) is 0 since the block starts from rest.

Initial potential energy ( $U_i$ ) due to spring compression is $\dfrac{1}{2} K x_i^2$ .

At a distance $x$ , the kinetic energy ( $K_f$ ) is $\dfrac{1}{2} mv^2$ and the potential energy ( $U_f$ ) is $\dfrac{1}{2} K x_f^2$ .

$K_i + U_i = K_f + U_f$ $0 + \dfrac{1}{2} K x_i^2 = \dfrac{1}{2} m v^2 + \dfrac{1}{2} K x_f^2$ Rearrange to find $v^2$ : $\dfrac{1}{2} m v^2 = \dfrac{1}{2} K (x_i^2 - x_f^2)$ Substitute known values: $\dfrac{1}{2} \times 2 \times v^2 = \dfrac{1}{2} \times 200 \times \left(1^2 - (2-x)^2 \right)$ Simplify and solve for $v^2$ : $v^2 = 100 \left[ 1 - (2-x)^2 \right]$ Finally, take the square root to find $v$ : $v = 10 \left[ 1 - (2-x)^2 \right]^{1/2}$ Therefore, the speed of the block at distance $x$ from the wall is given by the expression above.

Q63

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be :

A

{{\sqrt 3 } \over 2}

s

B

{3 \over 2}

s

C

{2 \over {\sqrt 3 }}

s

D

2\sqrt 3

s

Correct Answer

Option D

Solution

$\because$ g =

{{GM} \over {{R^2}}}

{{{g_p}} \over {{g_e}}}

=

{{{M_e}} \over {{M_e}}}{\left( {{{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}} \right)^2}

= 3

{\left( {{1 \over 3}} \right)^2}

=

{{1 \over 3}}

Also T $\propto$

{1 \over {\sqrt g }}

$\Rightarrow$

{{{T_p}} \over {{T_e}}}

=

\sqrt {{{{g_e}} \over {{g_p}}}}

=

\sqrt 3

$\Rightarrow$ Tp = 2

\sqrt 3

s

Q64

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by :

A 1 rad/s

B 10

-

3 rad/s

C 10

-

1 rad/s

D 10

-

5 rad/s

Correct Answer

Option B

Solution

Angular frequency of pendulum $\omega$ =

\sqrt {{{{g_{eff}}} \over \ell }}

$\therefore$

{{\Delta \omega } \over \omega }

=

{1 \over 2}

{{\Delta {g_{eff}}} \over {{g_{eff}}}}

\Delta

$\omega$ =

{1 \over 2}

{{\Delta g} \over g} \times \omega

[

{\omega _s}

= angular frequency of support]

\Delta

$\omega$ =

{1 \over 2} \times {{2A\omega _s^2} \over {100}} \times 100

\Delta \omega = {10^{ - 3}}

rad/sec.

Q65

In an engine the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :

A 0.1 Hz

B 1.2 Hz

C 0.7 Hz

D 1.9 Hz

Correct Answer

Option D

Solution

Here, Amplitude, A = 7 cm = 0.07 m When washer is no longer stays in contact with the piston, then the normal force on the washer is = 0 $\therefore$ Maximum acceleration of the washer, amax = $\omega$ 2A = g $\Rightarrow$ $\omega$ =

\sqrt {{g \over A}}

=

\sqrt {{{10} \over {0.07}}}

=

\sqrt {{{1000} \over 7}}

$\therefore$ Frequency of the piston, f =

{\omega \over {2\pi }}

=

{1 \over {2\pi }}\sqrt {{{1000} \over 7}}

= 1.9 Hz

Q66

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -

A

{{4\pi } \over 3}

B

{3 \over 8}\pi

C

{7 \over 3}\pi

D

{{8\pi } \over 3}

Correct Answer

Option D

Solution

v = \omega \sqrt {{A^2} - {x^2}} \,\,

. . .(1)

a = - {\omega ^2}x

. . .(2)

\left| v \right| = \left| a \right|

. . .(3)

\omega \sqrt {{A^2} - {x^2}} = {\omega ^2}x

{A^2} - {x^2} = {\omega ^2}{x^2}

{5^2} - {4^2} = {\omega ^2}\left( {{4^2}} \right)

\Rightarrow \,\,\,3 = \omega \times 4

T = 2\pi /\omega

Q67

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avogadro number = 6.02 × 1023 gm mole–1)

A 5.5 N/m

B 6.4 N/m

C 7.1 N/m

D 2.2 N/m

Correct Answer

Option C

Solution

6.02 $\times$ 1023 atoms of silver = 108 gm 1 atoms of silver =

{{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}

kg For a harmonic oscillator f =

{1 \over {2\pi }}

\sqrt {{k \over m}}

Where k = force constant $\Rightarrow$

\,\,\,

f2 =

{1 \over {4{\pi ^2}}}

\left( {{k \over m}} \right)

$\Rightarrow$

\,\,\,

k = mf2 $\times$ 4 $\pi$ 2 Given, f = 1012 m =

{{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}

\therefore\,\,\,

k =

{{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}

$\times$ 1012 $\times$ 4 $\pi$ 2 = 7.1 N/m

Q68

Two simple harmonic motions, as shown below, are at right angles. They are combined to form Lissajous figures. x(t) = A sin (at +

\delta

) y(t) = B sin (bt) Identify the correct match below.

A Parameters A

\ne

B, a = b;

\delta

= 0; Curve Parabola

B Parameters A = B, a = b;

\delta

=

{{\pi}/{2}}

Curve Line

C Parameters A

\ne

B, a = b;

\delta

=

{{\pi}/{2}}

Curve Ellipse

D Parameters A = B, a = 2b;

\delta

=

{{\pi}/{2}}

Curve Circle

Correct Answer

Option C

Solution

The given simple harmonic motions to form Lissajous figures are

x(t) = A\sin (at + \delta )

and

y(t) = B\sin (bt)

.

For parabola, conditions should be A = B or A $\ne$ B, a = 2b, $\delta$ = $\pi$ /2 For line, conditions should be A = B, a = b, $\delta$ = $\pi$ For circle, condition should be A = B, a = b; $\delta$ = $\pi$ /2 For ellipse, condition should be A $\ne$ B, a = b; $\delta$ = $\pi$ /2 Therefore, we obtain an ellipse.

Q69

The maximum velocity of a particle, executing simple harmonic motion with an amplitude

7

mm,

is

4.4

m/s.

The period of oscillation is

A

0.01

s

B

10

s

C

0.1

s

D

100

s

Correct Answer

Option A

Solution

Maximum velocity,

{v_{\max }} = a\omega ,\,\,\,\,\,{v_{\max }} = a \times {{2\pi } \over T}

\Rightarrow T = {{2\pi a} \over {{v_{\max }}}} = {{2 \times 3.14 \times 7 \times {{10}^{ - 3}}} \over {4.4}} \approx 0.01\,s

Q70

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and I, respectively. At time t = 0 one particle has displacement A while the other one has displacement

{{ - A} \over 2}

and they are moving towards each other. If they cross each other at time t, then t is :

A

{T \over 6}

B

{5T \over 6}

C

{T \over 3}

D

{T \over 4}

Correct Answer

Option A

Solution

Angular displacement ( $\theta$ 1) of particle 1. from equilibrium,

{y_1}

= A sin $\theta$ 1 $\Rightarrow$ A = Asin $\theta$ 1 $\Rightarrow$ sin $\theta$ 1 = 1 = sin

{\pi \over 2}

$\therefore$ $\theta$ 1 =

{\pi \over 2}

Similarly for particle 2 angular displacement $\theta$ 2 from equilibrium, y2 = Asin $\theta$ 2 $\Rightarrow$ $-$

{A \over 2}

= Asin $\theta$ 2 $\Rightarrow$ sin $\theta$ 2 = $-$

{1 \over 2}

= sin

\left( { - {\pi \over 3}} \right)

$\Rightarrow$ $\theta$ 2 = $-$

{{\pi \over 3}}

Relative angular displacement of the two particle, $\theta$ = $\theta$ 1 $-$ $\theta$ 2 =

{{\pi \over 2}}

$-$

\left( { - {\pi \over 6}} \right)

=

{{{2\pi } \over 3}}

Relative angular velocity $=$

\omega - \left( { - \omega } \right)

=

2\omega

If they cross each other at time t then, t =

{\theta \over {2\omega }}

=

{{2\pi } \over {3 \times 2\omega }}

=

{\pi \over {3 \times {{2\pi } \over T}}}

=

{T \over 6}