Oscillations Test

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Oscillations Test - 23

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Question 1
1 / -0

A desktop toy pendulum swings back and forth once every $$1.0 s$$. How long is this pendulum?

A
$$0.25\, m$$

B
$$0.50\, m$$

C
$$0.15\, m$$

D
$$0.30\, m$$

Solution

$$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$$ .. (1)

Putting $$T = 1$$ in eqn. (1),

We get $$1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$$ $$\Rightarrow\, L\, =\, \displaystyle \frac{g}{4\, \pi^2}\, =\, \displaystyle \frac{9.8}{4\, \times\, 3.14\, \times\, 3.14}m\, \Rightarrow\, L\, =\, \displaystyle \frac{9.8}{39.44}m\, =\, 0.2484\, m\, =\, 0.25\, m$$ (approx.)
Question 2
1 / -0

Calculate the period and frequency for the second hand, minute hand, and hour hand of a clock.

A
$$\mbox{For second hand, period = 1 min = 60 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 3600 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs = 43200 s; frequency = 2.3}\times10^{-5} Hz$$

B
$$\mbox{For second hand, period = 1 min = 360 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 600 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs =3200 s; frequency = 2.3}\times10^{-5} Hz$$

C
$$\mbox{For second hand, period = 1 min = 0 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 36 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs = 43200 s; frequency = 2.3}\times10^{-5} Hz$$

D
$$\mbox{For second hand, period = 1 min = 3660 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 3600 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs = 4300 s; frequency = 2.3}\times10^{-5} Hz$$

Solution

The frequency (f) of a wave is the number of full wave forms generated per second. This is the same as the number of repetitions per second or the number of oscillations per second.
Time Period (T) is the number of seconds per waveform, or the number of seconds per oscillation. It is clear that frequency and period are reciprocals.
For second hand, period = 1 min = 60 s; The frequency = $$\frac { 1 }{ t } =\frac { 1 }{ 60 } =0.0167 Hz$$.
For minute hand, period = 1 hr = 3600 s; The frequency = $$\frac { 1 }{ t } =\frac { 1 }{ 3600 } =2.77\times { 10 }^{ -4 }Hz$$.
For hour hand, period = 12 hrs = 43200 s; The frequency = $$\frac { 1 }{ t } =\frac { 1 }{ 43200 } =2.3\times { 10 }^{ -5 }Hz$$.
Question 3
1 / -0

Given equation of SHM is $$\displaystyle x=10\sin 10\pi t.$$ Find the distance between the two points where speed is $$\displaystyle 50\pi $$ cm/sec; x is in cm and t is in seconds.

A
$$5(\sqrt 3 +1)$$ cm

B
20 cm

C
17.32 cm

D
8.66 cm

Solution

$$x=10\,\sin10\pi t$$
$$\Rightarrow\cfrac{dx}{dt}=v=100\pi\cos(10\pi\,t)$$
$$\Rightarrow 50\pi=100\pi\cos(10\pi\,t)$$
$$\Rightarrow \cos(10\pi\,t)=\cfrac{1}{2}$$
$$\Rightarrow 10\pi\,t=\cfrac{\pi}{6},\cfrac{-\pi}{6}$$
So, $$ t=\cfrac{1}{60}sec$$
Now,
$$X=10\sin(10\pi\times\cfrac{1}{60})$$
$$\Rightarrow X=5\sqrt3\,cm$$
Also,
$$X=10\sin(10\pi(-\cfrac{1}{60}))$$
$$X=-5\,cm$$
$$ X = 5\ cm$$ (distance is always measured in +ve.)
$$Separation=[5\sqrt3-(-5)]\,cm$$
$$\Rightarrow\,Separation=5(\sqrt 3+1)\,cm$$
Question 4
1 / -0

For a particle performing SHM, equation of motion is given as $$\displaystyle \frac{d^{2}x}{dt^{2}}+4x=0$$. Find the time period

A
$$\displaystyle T=\pi $$

B
$$\displaystyle T=2\pi $$

C
$$\displaystyle T=3\pi $$

D
$$\displaystyle T=4\pi $$

Solution

$$\displaystyle \frac{d^{2}x}{dt^{2}}=-4x$$ $$\displaystyle \omega ^{2}=4$$ $$\displaystyle \omega=2$$
Time period $$\displaystyle T=\frac{2\pi }{\omega }=\pi $$
Question 5
1 / -0

The bob of mass $$50 gms$$ of a simple pendulum is replaced by another bob of mass $$75 gms$$. The time period of the simple pendulum

A
Becomes $${75}/{50}$$ times the original period

B
Becomes $${50}/{75}$$ times the original period

C
Becomes $$125$$ times the original period

D
Remains unchanged

Solution

As the periodic time of a pendulum does not depend on the mass of the bob, so on replacing the bob of mass $$50 gm$$ by another bob of mass $$75 gm$$ the time period of the simple pendulum will remain unchanged.
Question 6
1 / -0

A hollow pendulum bob filled with water has a small hole at the bottom through which water escapes at a constant rate. Which of the following statements describes the variation of the time period (T) of the pendulum as the water flows out?

A
T decreases first and then increases.

B
T increases first and then decreases.

C
T increases throughout.

D
T does not change.

Solution

$$\displaystyle T = 2\pi \sqrt{\frac{l}{g}}$$
First distance of comfrom suspension point will increase then decrease.
Question 7
1 / -0

The magnitude of acceleration of particle executing SHM at the position of maximum displacement is:

A
zero

B
minimum

C
maximum

D
none of these

Solution

The acceleration of the particle executing S.H.M is, $$a = -\omega^2 y.$$
At maximum displacement (at extreme position), $$y = \pm a$$
Hence $$|a_{max}|= \omega^2a$$. i.e., acceleration will be maximum.
Question 8
1 / -0

Directions For Questions

A particle of mass 2 kg is acted upon by a force of F = (8 - 2x) N. It is released from rest at x = 6 m.

...view full instructions

The equilibrium position of the particle is

A
x = 4

B
x = 6

C
x = 2

D
x = 3

Solution

F = 8 - 2x or F = -2(x - 4) at equilibrium position F = 0 $$\Rightarrow$$ x = 4 is equilibrium position
Hence the motion of particle is SHM with force constant 2 and equilibrium position x = 4
Equilibrium position is x = 4
Question 9
1 / -0

The bob of a second's pendulum is replaced by another bob of double mass. The new time period will be

A
4 sec

B
1 sec

C
2 sec

D
3 sec

Solution
Question 10
1 / -0

If the period of oscillation of a simple pendulum is 4 second and we want to convert it into a second pendulum, then we have to

A
Make the length of the pendulum one fourth of the previous length

B
Double the length of the pendulum

C
Make the length of the pendulum half of the previous length

D
Double the mass of the bob

Solution

The time period of a simple pendulum is proportional to the square root of the length i.e., $$T \infty \sqrt{l}$$. To convert 4 second to 2 we have to make the length of the pendulum one fourth of the previous length.

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Oscillations Test - 23

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Oscillations Test - 23

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Question 1

$$0.25\, m$$

$$0.50\, m$$

$$0.15\, m$$

$$0.30\, m$$

Solution

Question 2

$$\mbox{For second hand, period = 1 min = 60 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 3600 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs = 43200 s; frequency = 2.3}\times10^{-5} Hz$$

$$\mbox{For second hand, period = 1 min = 360 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 600 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs =3200 s; frequency = 2.3}\times10^{-5} Hz$$

$$\mbox{For second hand, period = 1 min = 0 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 36 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs = 43200 s; frequency = 2.3}\times10^{-5} Hz$$

$$\mbox{For second hand, period = 1 min = 3660 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 3600 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs = 4300 s; frequency = 2.3}\times10^{-5} Hz$$

Solution

Question 3

$$5(\sqrt 3 +1)$$ cm

20 cm

17.32 cm

8.66 cm

Solution

Question 4

$$\displaystyle T=\pi $$

$$\displaystyle T=2\pi $$

$$\displaystyle T=3\pi $$

$$\displaystyle T=4\pi $$

Solution

Question 5

Becomes $${75}/{50}$$ times the original period

Becomes $${50}/{75}$$ times the original period

Becomes $$125$$ times the original period

Remains unchanged

Solution

Question 6

T decreases first and then increases.

T increases first and then decreases.

T increases throughout.

T does not change.

Solution

Question 7

zero

minimum

maximum

none of these

Solution

Question 8

x = 4

x = 6

x = 2

x = 3

Solution

Question 9

4 sec

1 sec

2 sec

3 sec

Solution

Question 10

Make the length of the pendulum one fourth of the previous length

Double the length of the pendulum

Make the length of the pendulum half of the previous length

Double the mass of the bob

Solution

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