Oscillations Test

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Oscillations Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $$10\ cm$$ to $$8\ cm$$ in $$40$$ seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $$1.3$$, the time In which amplitude of this pendulum will reduce from $$10\ cm$$ to $$5\ cm$$ in carbon dioxide will be close to (ln $$5=1.601, \ln { 2 } 2=0.693$$)

A
$$231\ s$$

B
$$208\ s$$

C
$$161\ s$$

D
$$142\ s$$

Solution

$$\textbf{Solution}:$$
We know that the amplitude of damped oscillation is
$$A=A_0 e^{\dfrac{-bt}{2m}}$$

$$\implies 8=10e^{(-\dfrac{b_{air} \times 40}{2m})}$$

and
$$5=10e^{-(\dfrac{b_{co_2} \times t_2}{2m})}$$

$$\implies \dfrac{In\dfrac{5}{4}}{In2}=\dfrac{b_{air} \times 40}{b_{co_2} \times t_2}$$

Therefore ,

$$t_2=\dfrac{b_{air}}{b_{co}} \times \dfrac{40 \times In2}{In\dfrac{5}{4}}=161s$$

$$\textbf{Hence C is the correct option}$$
Question 2
1 / -0

The potential energy of a particle of mass 10g varies as its displacement from its mean position given by $$U = 3x^2 + 3$$, then, the particle performs a

A
SHM with time period $$T = \pi$$ s

B
Linear motion

C
SHM with time period $$T =\dfrac{ \pi}{5\sqrt{6}}$$ s

D
SHM with time period $$T = \pi/5$$ s

Solution

The force acting on the system is $$F = -dU/dx = -6x \implies a=-(6/0.01)x=-600x$$. The angular velocity of the particle is $$\omega= \sqrt{600}$$

Thus, the motion of the particle is SHM with time period $$T=2 \pi/\sqrt{600} $$ or $$T = \pi/5\sqrt{6}$$ s

The correct option is (c)
Question 3
1 / -0

A particle executes SHM along x axis and is at the mean position at t=0. What is its velocity at its mean position. The amplitude of SHM is 5 cm and angular frequency is 2 rad/s:

A
5 cm/s

B
10 cm/s

C
7 cm/s

D
0 cm/s

Solution

The particle has its maximum velocity at mean position and is given by $$v=A \omega = 10 $$ cm/s

The correct option is (b)
Question 4
1 / -0

A particle executes SHM given by the equation $$x = 4 sin (2 \pi t +\pi/4)$$, what will be the velocity of the particle at t = (1/8)th sec;

A
velocity = 0

B
velocity = maximum

C
velocity = minimum

D
The particle's speed cannot be determined

Solution

$$x = 4 sin (2 \pi t +\pi/4)$$
Differentiating x w.r.t. t, we get, $$dx/dt =v = 8 \pi cos (2 \pi t +\pi/4)$$
Substituting t = 1/8, we get, $$v=8 \pi cos (2 \pi t +\pi/4)= cos (\pi/2)=0$$

At t = 1/8th of a sec, the particle's velocity will be zero . Thus we infer that the particle is at the extreme position

The correct option is (a)
Question 5
1 / -0

The potential energy of a particle is directly proportional to its linear displacement from its mean position. Then, the particle performs a

A
Retarding straight line motion

B
Damped SHM

C
Linear SHM

D
Angular SHM

Solution

Potential energy U = Kx. Thus the force acting is = -dU/dx = -K

The force is a retarding constant force and the motion of the particle will be in a straight line

The correct option is (a)
Question 6
1 / -0

At what position along a straight line will the velocity be zero for a particle executing SHM

A
Mean position

B
$$x=A/2$$

C
Extremes

D
$$x= - A/2$$

Solution

The velocity and displacement follows an equation given by $$v^2=\omega (A^2-x^2)$$. If v has to be zero, then x=A.

If the particle was at the extreme, the particle will have a zero velocity

Thus the correct option is option(c)
Question 7
1 / -0

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that is at a distance $$\dfrac{2A}{3}$$ from equilibrium position. The new amplitude of the motion is:

A
$$3A$$

B
$$A\sqrt3$$

C
$$\dfrac{7A}{3}$$

D
$$\dfrac{A}{3}\sqrt{41}$$

Solution

$$3\omega \sqrt {A^2-\left(\dfrac{2A}{3}\right)^2}=\omega \sqrt{A_1^2-\left(\dfrac{2A}{3}\right)^2}$$
$$\therefore A_1=\dfrac{7A}{3}$$
Question 8
1 / -0

What type of curve do we get, if $$x^2$$ and $$v^2$$ are plotted for a particle executing SHM, x and v are the position and velocity of the particle:

A
Circle

B
Straight line

C
Ellipse

D
Rectangle

Solution

The equation of SHM, when the particle started at the mean position at t = 0 is given by $$x=A sin (\omega t )$$

Differentiating x w.r.t. t, we get $$v = A \omega cos \omega t $$

To eliminate $$\theta$$, we can do
$$(x/A)^2 + (v/A \omega)=1 \implies $$ , the plot will be a circle

The correct option is (c)
Question 9
1 / -0

A particle executes SHM from its mean position at t=0 with an amplitude A, what will be its velocity at x=A/2, in its forward motion towards the extreme:

A
$$\sqrt(3)A \omega/2$$

B
$$\sqrt(3)A \omega/4$$

C
$$A \omega/2$$

D
$$A \omega$$

Solution

The equation of SHM, when the particle started at the mean position at t = 0 is given by $$x=A sin (\omega t )$$

At x=(A/2), the particle reaches it at time t given by $$sin \omega t=0.5. \implies \omega t = \pi/6$$.

Differentiating x w.r.t. t, we get $$v = A \omega cos \omega t. $$. Substituting the value of $$\omega t$$, we get $$v =A \omega cos 30 = \sqrt(3)A \omega/2$$

The correct option is (a)
Question 10
1 / -0

The time period of oscillation of magnet in a vibrating magnetometer is $$1.5$$ sec. The time period of oscillation of another magnet similar in size and mass but having one-fourth the magnetic moment than that of the first magnet oscillating at the same place will be

A
$$0.75$$ sec

B
$$1.5$$ sec

C
$$3.0$$ sec

D
$$6.0$$ sec

Solution

$$\begin{array}{l} \dfrac { { { T_{ 1 } } } }{ { { T_{ 2 } } } } =\sqrt { \dfrac { { { M_{ 2 } } } }{ { { M_{ 1 } } } } } =\sqrt { \dfrac { { { M_{ 1 } } } }{ { \dfrac { 4 }{ { { M_{ 1 } } } } } } } =\dfrac { { { T_{ 1 } } } }{ { { T_{ 2 } } } } =\dfrac { 1 }{ 2 } \\ \Rightarrow { T_{ 2 } }=3 \end{array}$$

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Re-Attempt Weekly Quiz Competition

Oscillations Test - 35

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Oscillations Test - 35

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SHARING IS CARING

Question 1

$$231\ s$$

$$208\ s$$

$$161\ s$$

$$142\ s$$

Solution

Question 2

SHM with time period $$T = \pi$$ s

Linear motion

SHM with time period $$T =\dfrac{ \pi}{5\sqrt{6}}$$ s

SHM with time period $$T = \pi/5$$ s

Solution

Question 3

5 cm/s

10 cm/s

7 cm/s

0 cm/s

Solution

Question 4

velocity = 0

velocity = maximum

velocity = minimum

The particle's speed cannot be determined

Solution

Question 5

Retarding straight line motion

Damped SHM

Linear SHM

Angular SHM

Solution

Question 6

Mean position

$$x=A/2$$

Extremes

$$x= - A/2$$

Solution

Question 7

$$3A$$

$$A\sqrt3$$

$$\dfrac{7A}{3}$$

$$\dfrac{A}{3}\sqrt{41}$$

Solution

Question 8

Circle

Straight line

Ellipse

Rectangle

Solution

Question 9

$$\sqrt(3)A \omega/2$$

$$\sqrt(3)A \omega/4$$

$$A \omega/2$$

$$A \omega$$

Solution

Question 10

$$0.75$$ sec

$$1.5$$ sec

$$3.0$$ sec

$$6.0$$ sec

Solution

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