Oscillations Test

Result

Oscillations Test - 44

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 oscillating in simple harmonic motion is :

A
Never in equilibrium because it is in motion

B
Never in equilibrium because there is always a force

C
In equilibrium at the ends of its path because of zero velocity there

D
In equilibrium at the centre of its path because the acceleration is zero there

Solution
Question 2
1 / -0

A particle is executing $$S.H.M.$$ If the amplitude is $$2m$$ and periodic time $$2 \ \ seconds$$, then the max. velocity of the particle will be

A
$$\pi m/s$$

B
$$\sqrt 2 \pi m/s$$

C
$$2 \pi m/s$$

D
$$4 \pi m/s$$

Solution

In an $$S.H.M$$
$$A=2\ m, T=2\ s$$
So, $$\cfrac{2\pi}{\omega}=2\implies \omega=\pi$$
$$V_{max}=A\omega$$
$$=2\pi\ m/s$$
Question 3
1 / -0

A particle of mass $$2$$kg, executing SHM has amplitude $$20$$cm and time period $$1$$s. Its maximum speed is

A
$$0.314$$m/s

B
$$0.628$$m/s

C
$$1.256$$m/s

D
$$2.512$$m/s

Solution

$$v_{max}=Aw$$

$$=20cm \times \dfrac{2\pi}{T}$$

$$=0.2m \times \dfrac{2\pi}{1s}$$

$$=1.256m/s$$
Question 4
1 / -0

A spring-block system undergoes simple harmonic motion on a smooth horizontal surface. The block is now given some positive charge, and a uniform horizontal electric field to the right is switched on. As a result,

A
the time period of oscillation will increase

B
the time period of oscillation will decrease

C
the time period of oscillation will remain unaffected

D
the mean position of simple harmonic motion will shift to the left

Solution

When a constant force is superimposed on a system which undergoes SHM along the line of SHM, the time period does not change as it depends on mass of the block and force constant of spring.
The mean position changes as this is the point where net force on the particle is zero.
Question 5
1 / -0

A body of mass $$100 gm$$ is performing $$S.H.M$$ with a period of $$4.5s$$ and amplitude $$7 cm$$ . Velocity of the body at equilibrium position in $$cm/s$$ is

A
$$10$$

B
$$8$$

C
$$6$$

D
$$5$$

Solution
Question 6
1 / -0

$$\upsilon $$ as shown in figure. The velocity of the block on the plane 'u' is

A
$$\upsilon$$ cos $$ec \theta$$

B
$$\upsilon$$ sin $$ \theta$$

C
$$\upsilon$$ cos $$ \theta$$

D
$$\upsilon$$ sec $$ \theta$$
Question 7
1 / -0

A particle vibrates in $$SHM$$ along a straight line. Its greatest acceleration is $$5 \pi ^ { 2 } \: { cms } ^ { - 2 }$$ and when its distance from the equilibrium position is $$4\ cm$$, the velocity of the particle is $$3 \pi \:{ cms } ^ { - 1 }$$.

A
the amplitude is $$10\ cm$$

B
the period of oscillation $$2\ sec$$

C
the amplitude is $$0.5 \ cm$$

D
the period of oscillation $$4\ sec$$

Solution
Question 8
1 / -0

The acceleration of a particle in $$SHM$$ at $$5\ cm$$ from its mean position is 20 $$cm/{ sec }^{ 2 }$$. The value of angular frequency in radians/ sec will be :

A
$$2$$

B
$$4$$

C
$$10$$

D
$$14$$

Solution
Question 9
1 / -0

A particle has simple harmonic motion. The equa-tion of its motion is $$x=5\sin (4t- \pi / 6)$$, where $$x$$ is its displacement. If the displacement of the particle is $$3\ units$$, then its velocity is

A
$$2 \pi /3$$ units

B
$$5 \pi /6$$ units

C
$$20$$ units

D
$$16$$ units

Solution

From the given equation,
The amplitude of the wave is $$a=5$$
and the angular velocity of the wave is $$\omega =4$$

The velocity of the wave can be obtained as:
$$\therefore v=\omega \sqrt{a^2-y^2}$$

$$=4\sqrt{(5)^2-(3)^2}=16\ units$$
Question 10
1 / -0

The amplitude and time period of a particle performing $$S.H.M$$ are ;a;and $$T$$ respectively. The displacement at which its velocity will be half the max. velocity will be

A
$$\dfrac{a}{2}$$

B
$$\dfrac{a}{3}$$

C
$$\sqrt 3 {\dfrac{a}{2}}$$

D
$$\dfrac{2a}{\sqrt 3}$$

Solution

The displacement of a point $$\times$$ subjected to sumis $$x=asinT$$
The velocity $$V$$ of that point $$V=acosT$$
When $$V=\dfrac { acosT }{ 2 } $$
$$\Rightarrow cos\left( 0.5 \right) =600$$
At this velocity $$\left( V={ 60 }^{ 0 } \right) $$ Displacement
$$x=a\sin{ 60 }^{ 0 }=0.866a$$