Oscillations Test

Result

Oscillations Test - 28

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A particle vibrating simple harmonically has an acceleration of $$16{cms}^{-2}$$ when it is at a distance of $$4cm$$ from the mean position. Its time period is

A
$$1s$$

B
$$2.572s$$

C
$$3.142s$$

D
$$6.028s$$

Solution

The equation of motion for an SHM is given as
$$x=Asin(\omega t+\phi)$$
$$\implies a=\dfrac{d^2x}{dt^2}$$
$$=-A\omega^2sin(\omega t+\phi)$$
$$=-\omega^2 x$$
Hence for the given position,
$$16=\omega ^2 (4)$$
$$\implies \omega=2$$
Hence the time period of the motion is $$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{2}=\pi\approx 3.142s$$
Question 2
1 / -0

The simple harmonic motion of a particle is given by $$x=a\sin 2\pi t$$. Then, the location of the particle from its mean position at a time $$1/8$$th of a second is:

A
$$a$$

B
$$\displaystyle\frac{a}{2}$$

C
$$\displaystyle\frac{a}{\sqrt{2}}$$

D
$$\displaystyle\frac{a}{4}$$

E
$$\displaystyle\frac{a}{8}$$

Solution

Position of particle executing SHM from mean position $$x=a\sin 2\pi t$$
$$x=a\sin 2\pi\times \displaystyle\frac{1}{8}$$
$$x=a\sin \displaystyle\frac{\pi}{4}$$
$$x=\displaystyle\frac{a}{\sqrt{2}}$$.
Question 3
1 / -0

A particle which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance $$x$$ of the particle from the origin is $$F(x)=-kx+a{ x }^{ 3 }$$. Here, $$k$$ and $$a$$ are positive constants. For $$x\ge 0$$, the functional form of the potential energy $${U}_{(x)}$$ of the particle is :

A

B

C

D

Solution

We know $$F=-\cfrac { dU }{ dx } $$
$$\Rightarrow dU=-F.dx$$
$$\Rightarrow U=-\int _{ 0 }^{ x }{ \left( -kx+a{ x }^{ 3 } \right) } dx\Rightarrow U=\cfrac { k{ x }^{ 2 } }{ 2 } -\cfrac { a{ x }^{ 4 } }{ 4 } \quad $$
$$\therefore$$ we get $$U=0$$ at $$x=0$$ and $$x=\sqrt { \cfrac { 2k }{ a } } $$
Also we get $$U=$$ negative for $$x> \sqrt { \cfrac { 2k }{ a } } $$
From the given function we can see that $$F=0$$ at $$x=0$$ i.e slope of U-x graph is zero at $$x=0$$
Question 4
1 / -0

Which one of the following is simple harmonic?

A
Rotation of earth around the sun

B
Rotation of earth about its own axis

C
Revolving motion of a top

D
Motion of a steel ball in a viscous medium

E
Motion of oscillating liquid column in U-tube

Solution

Simple harmonic motion is a special type of periodic motion, in which a particle moves to and fro repeatedly about a mean position. So, motion of oscillating liquid column in U-tube is simple harmonic.
Question 5
1 / -0

A 2 kg block is dropped from a height $$ 0.4 m $$ on a spring of force constant $$k=1960 N/m$$.The maximum compression of spring is

A
$$ 0.1 m$$

B
$$ 0.2 m$$

C
$$ 0.3 m$$

D
$$ 0.4 m$$

Solution

By law of conservation of energy , we have loss in gravitational potential energy=gain in elastic potential energy
$$\Rightarrow mg(h+x)=\dfrac{1}{2}kx^{2}$$
$$2\times 10(0.4+x)=\dfrac{1}{2}\times 1960\times x^{2}$$
$$\Rightarrow 8+20x=980\,x^{2}$$
$$\Rightarrow 980x^{2}-20x-8=0$$
Solving ,we get
$$x=0.1 m$$
Question 6
1 / -0

Four springs having the same force constant and a mass $$M$$ are connected between rigid walls as shown in figure. The system with smallest time period is

A

B

C

D

Solution

$$A)$$Time period $${ T }_{ A }=2\pi \sqrt { \cfrac { m }{ 2k } } $$
$$B)$$ Time period, $${ T }_{ B }=2\pi \sqrt { \cfrac { m }{ k+\cfrac { k }{ 3 } } } $$
$$C)$$ Time period, $${ T }_{ C }=2\pi \sqrt { \cfrac { 4m }{ k } } =4\pi \sqrt { \cfrac { m }{ k } } $$
$$D)$$ Time period, $${ T }_{ D }=2\pi \sqrt { \cfrac { m }{ 4k } } =\pi \sqrt { \cfrac { m }{ k } } $$
Question 7
1 / -0

Velocity at mean position of a particle executing SHM is v, then velocity of the particle at a distance equal to half of the amplitude is

A
$$\frac {\sqrt{3}}{2}v$$

B
$$\frac {2} {\sqrt{3}}v$$

C
$$2v$$

D
$$v$$

Solution

At mean position $$v = a\omega$$
Velocity at half the amplitude is
$$v'= \omega \sqrt {a^2-y^2}$$
$$=\omega \sqrt{a^2-\frac{a^2}{4}}= \frac{\sqrt{3}}{2}a\omega = \frac{\sqrt{3}}{2} v$$
Question 8
1 / -0

A spring with force constant $$k$$ is initially stretched by $$x_{1}$$. If it further stretched by $$x_{2}$$, then the increase in its potential energy is

A
$$\dfrac {1}{2}k(x_{2} - x_{1})^{2}$$

B
$$\dfrac {1}{2}kx_{2}(x_{2} + 2x_{1})$$

C
$$\dfrac {1}{2}\ kx_{1}^{2} + \dfrac {1}{2}\ kx_{2}^{2}$$

D
$$\dfrac {1}{2}\ k(x_{1} + x_{2})^{2}$$

E
$$\dfrac {1}{2}\ k(x_{1} + x_{2})^{3}$$

Solution

According to the question, in the first condition
$$U_{1} = \dfrac {1}{2}kx_{1}^{2} .... (i)$$
In the second condition,
$$U_{2} = \dfrac {1}{2}k (x_{1} + x_{2})^{2} .... (ii)$$
Increment in the potential energy
$$= U_{2} - U_{1}$$
$$= \dfrac {1}{2} k(x_{1}^{2} + x_{2}^{2} + 2x_{1} x_{2}) - \dfrac {1}{2} kx_{1}^{2}$$
$$= \dfrac {1}{2} k(x_{1}^{2} + x_{2}^{2} + 2x_{1}x_{2} - x_{1}^{2})$$
$$= \dfrac {1}{2} k[2x_{1}x_{2} + x_{2}^{2}]$$
$$= \dfrac {1}{2} kx_{2} (x_{1} + 2x_{1})$$.
Question 9
1 / -0

The displacement of a particle in SHM is $$x=10\sin \begin{pmatrix}2t-\dfrac{\pi}{6}\end{pmatrix}m.$$ When its displacement is 6 m, the velocity of the particle (in $$ms^{-1}$$) is

A
8

B
24

C
16

D
10

E
12

Solution

Given, $$x=10\sin(2t-\pi/6)$$
Putting $$x=6 \Rightarrow 6=10\sin(2t-\pi/6)$$
So, $$\sin(2t-\pi/6)=3/5$$
Using formula, $$\sin^2\theta+cos^2\theta=1$$
$$cos(2t-\pi/6)=\sqrt{1-(3/5)^2}=4/5$$
Now, velocity $$v=\dfrac{dx}{dt}=20cos(2t-\pi/6)=20\times (4/5)=16 \, m/s$$
Question 10
1 / -0

Which one of the following represents simple harmonic motion ?

A
$$x^{2}=a+bv$$

B
$$x=\sqrt{a+bv^{2}}$$

C
$$x=a-bv$$

D
$$x=\sqrt{a-bv^{2}}$$

Solution