Oscillations Test

Result

Oscillations Test - 47

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

Two particles of same period $$(T)$$ and amplitude undergo $$SHM$$ along the same line with initial phase of $$\pi/6$$. If they start at the same instant and at same point along opposite directions, find the time after which they will meet again for the first time:

A
$$3T/8$$

B
$$T/4$$

C
$$T/2$$

D
$$T$$

Solution

$$x=A\sin \left ( wt-\dfrac{\pi}{2} \right )$$

$$\sin \left ( wt-\dfrac{\pi}{2} \right )=\sin wt$$

$$2wt-\dfrac{\pi}{2}=\pi $$

$$2wt=\dfrac{3}{2}\pi $$

$$wt=\dfrac{3}{4}\pi $$

$$t=\dfrac{3}{4}\dfrac{\pi}{w} $$

$$t=\dfrac{2\pi}{w}\left ( \dfrac{3}{8} \right )$$

$$\therefore t=\dfrac{3T}{8}$$
Question 2
1 / -0

The total energy of a particle executing SHM is directly proportional to the square of the following quantity.

A
Acceleration

B
Amplitude

C
Time period

D
Mass

Solution
Question 3
1 / -0

A particle executes SHM according to equation $$x = 10 \cos[2\pi t+\pi / 2]$$, where $$t$$ is in second. The magnitude of the velocity of the particle at $$t=1/6s$$ will be:

A
$$2.47$$

B
$$2.86$$

C
$$2.05$$

D
$$31.4$$

Solution
Question 4
1 / -0

A particle executes $$SHM$$ and its position with time as $$x = A \sin\omega t$$. Its average speed during its motion from mean position to mid-point of mean and extreme position is

A
$$Zero$$

B
$$\cfrac{3A\omega}{\pi}$$

C
$$\cfrac{A\omega}{2\pi}$$

D
$$\cfrac{2A\omega}{\pi}$$

Solution
Question 5
1 / -0

A student says that he had applied a force $$F= -k\sqrt {x}$$ on a particle and the particle moves in simple harmonic motion. He refuses to tell whether $$k$$ is a constant or not. Assume that he has worked only with positive $$x$$ and no other force acted on the particle.

A
As $$x$$ increases $$k$$ increases

B
As $$x$$ increases $$k$$ decreases

C
As $$x$$ increases $$k$$ remains constant

D
The motion cannot be simple harmonic

Solution

Correct Answer: Option A
Step 1: Use force law for SHM.
As the motion of the particle is Simple Harmonic Motion.
From force law,
$$F = -kx$$
where $$k$$ is a constant $$= m\omega^2$$
$$F=-m\omega^2 x$$ ....... $$(I)$$

Force applied by the student is
$$F= -k \sqrt x $$ ....... $$(II)$$
So, equating $$(I)$$ and $$(II)$$

$$ -k \sqrt x = - m\omega^2 x$$

$$k=m\omega^2 \sqrt x$$
so,
$$k \propto \sqrt x$$

As $$x$$ increases, $$ \sqrt x$$ increases and thus $$k$$ increases.

Hence, the correct answer is option A.
Question 6
1 / -0

A particle undergoes $$SHM$$ of period $$T$$. The time taken to complete $$3/8th$$ oscillation starting from the mean position is

A
$$\dfrac{3}{8}T$$

B
$$\dfrac{5}{8}T$$

C
$$\dfrac{5}{12}T$$

D
$$\dfrac{7}{12}T$$

Solution

Time to complete $$\dfrac{1}{8}$$th oscillation from extreme position is,

$$y=\dfrac{a}{2}=a\cos wt=a \cos \dfrac{2\pi}{T}t$$

$$\Rightarrow \cos \dfrac{\pi}{3}=\cos \dfrac{2\pi t}{T}$$

$$\Rightarrow t=\dfrac{T}{6}$$

w.k.t time taken to complete one fourth oscillation is $$\dfrac{T}{4}$$

Therefore the time to complete $$\dfrac{3}{8}$$th oscillation

$$=\dfrac{T}{4}+\dfrac{T}{6}$$

$$=\dfrac{5T}{12}$$
Question 7
1 / -0

The mass of particle executing S.H.M is 1 gm.If its periodic time is $$\pi $$ seconds, the value of force constant is:-

A
4 dynes/cm

B
4 N/cm

C
4 N/m

D
4 dynes/m

Solution

Given, mass $$m=1 gm$$; time period $$T=\pi$$ sec
The angular frequency $$\omega=\frac{2\pi}{T}=\frac{2\pi}{\pi}=2$$
As $$\omega=\sqrt{k/m}$$ where ($$k$$ is the force constant),
$$k=m\omega^2=(1 gm)(2)^2=4$$ dynes/cm
Question 8
1 / -0

A particle is executing SHM about y=0 along y-axis. Its position at an instant is given by y=(7m) sin $$\left( \pi f \right) .$$ its average velocity for a time interval 0 to 0.5 s is

A
14 m/s

B
7 m/s

C
$$\dfrac { 1 }{ 7 } m/s$$

D
28 m/s

Solution
Question 9
1 / -0

A small body of mass $$0.10 kg$$ is executing S.H.M of amplitude $$1.0m$$ and period $$0.20sec$$. The maximum force acting on it is:

A
$$98.596N$$

B
$$985.96N$$

C
$$100.2N$$

D
$$76.23N$$

Solution

Given:
The mass of the body is $$m=0.10\ kg$$
The amplitude of SHM $$a=1.0\ m$$.
The time period of oscillation is $$T=0.20\ s$$

Maximum acceleration of the syustem is given by:
$$A_{max}=a \omega^2$$ $$\because \omega=\dfrac{2\pi}{T}$$

$$=\dfrac{a \times 4 \pi^2}{T^2}=\dfrac{1 \times 4 \times (3.14)^2}{0.2 \times 0.2}$$

So, the maximum force acting on the body will be:
$$F_{max}= m \times A_{max}=\dfrac{0.1 \times 4 \times (3.14)^2}{0.2 \times 0.2}=98.596N$$
Question 10
1 / -0

The time period of SHM of a particle is $$12 s.$$ The phase difference between the position at $$t =3s,$$ and $$t = 4s$$ will be :

A
$$\dfrac{\pi }{4}$$

B
$$\dfrac{3\pi }{5}$$

C
$$\dfrac{\pi }{6}$$

D
$$\dfrac{\pi }{2}$$

Solution