Oscillations Test

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Oscillations Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

A thin rectangular magnet suspended freely has a period of oscillation $$4\ s$$. If it is broken into two halves each having half their initial length, then when suspended similarly, the time period of oscillation of each part will be

A
$$4\ s$$

B
$$2\ s$$

C
$$1\ s$$

D
$$4\sqrt{2}\ s$$

Solution

A thin rectangular magnet suspended freely has a period of oscillation $$T=4s$$
We know that
$$T=2\pi\sqrt{\cfrac{l}{g}}\\ \cfrac{T_1}{T_2}=\sqrt{\cfrac{l_1}{l_2}}\\ \cfrac{4}{T_2}=\sqrt{\cfrac{l_1}{l_1/2}}=\sqrt2\quad or\quad T_2=4\sqrt2s$$
Question 2
1 / -0

The displacement of a particle in S.H.M. is indicated by equation $$y\ = \ 10\ sin(20t\ +\ \pi/3)$$ where $$y$$ is in metres. The value of time period of vibration will be (in seconds):

A
$$10/\pi $$

B
$$\pi /10$$

C
$$2\pi /10$$

D
$$10/\pi 2$$

Solution

$$y=10\sin(20t+\dfrac{\pi}{3})$$
Comparing with below equation,
$$y=A\sin(\omega t+\phi)$$ where $$\omega=$$angular velocity=20
Time period $$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\omega}=\dfrac{\pi}{10}$$
Question 3
1 / -0

For a particle $$SHM$$, equation of motion is given as $$\dfrac{d^{2}x}{dt^{2}}+4x=0$$. The time period of this wave is

A
$$2 \pi$$

B
$$\pi$$

C
$$\dfrac{2 \pi}{3}$$

D
$$\dfrac{\pi}{3}$$

Solution
Question 4
1 / -0

Two identical bar magnets are placed on above the other such that they are mutually perpendicular and bisect each other. The time period of this combination in a horizontal magnetic filed is T. The time period of each magnet in the same field is

A
$$\sqrt{2}T$$

B
$$\sqrt[4]2T$$

C
$$\dfrac 1{\sqrt[4]2}T$$

D
$$\dfrac 1{\sqrt2}T$$

Solution
Question 5
1 / -0

A particle is executing simple harmonic motion between extreme positions given by $$(-1.-2.-3)cm$$ and $$(1,2,1)cm$$. Its amplitude of oscillation is :

A
$$6cm$$

B
$$4cm$$

C
$$2cm$$

D
$$3cm$$

Solution

Let $$P(-1, -2, -3)\ cm$$ and $$Q(1, 2, 1)\ cm$$ are the points of extreme positions in S.H.M.
Then $$2 \times A=d(P, Q)$$
$$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

$$=\sqrt{(1-(-1))^2+(2-(-2))^2+(1-(-3))^2}$$

$$=\sqrt{2^2+4^2+4^2}$$

$$2A=\sqrt{36}$$

Amplitude $$=\dfrac{6}{2}=3 cm$$
Question 6
1 / -0

A body is executing S.H.M. When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocity of the body is 10 cm/sec and 8 cm/sec. Then the time period of the body is:

A
$$2\pi \sec $$

B
$$\pi /2\sec $$

C
$$\;\pi \sec \;$$

D
$$3\pi /2\sec $$

Solution

When the displacement from the mean position is $$4\ cm$$
Potential energy=$$\cfrac{1}{2} 4^2u\\=8u\\=8(m\omega^2)$$
Kinetic energy$$=\cfrac{1}{2}mv^2\\=\cfrac{1}{2}10^2m\\=50m$$
Total energy$$=8m\omega^2+50m$$
When the displacement from the mean position is $$5\ cm$$
Potential energy=$$\cfrac{1}{2} 5^2u\\=12.5u\\=12.5(m\omega^2)$$
Kinetic energy$$=\cfrac{1}{2}mv^2\\=\cfrac{1}{2}8^2m\\320m$$
Total energy$$=12.5m\omega^2+32m$$
Now as per conservation of energy
$$=8m\omega^2+50m$$$$=12.5m\omega^2+32m$$
$$\implies \omega=2$$
And $$\omega=\cfrac{2\pi}{T}=2\implies T=\pi\ s$$
Question 7
1 / -0

The acceleration of a particle in SHM at 5 cm from the mean position of $$20 cm/sec^2$$. The value of angular frequency in radians per second will be

A
2

B
4

C
10

D
14

Solution

$$\textbf{Step 1 - Acceleration of particle}$$
Acceleration of particle at distance $$x$$ from mean position, $$a = -\omega^{2}x$$
Given
$$a = -20\ cm/\sec^{2}$$ [acceleration is towards mean position always so taken negative]
$$-20 = -\omega^{2}x$$ [here $$\omega$$ is angular frequency]
$$-20 cm/\sec^{2} = -\omega^{2} \times 5\ cm$$
$$\omega = 2\ rad/\sec$$

Hence option (A) correct.
Question 8
1 / -0

A particle executing S.H.M. completes a distance (taking friction as negligible) in one complete one time period.

A
Four times the amplitude

B
Two times the amplitude

C
One times the amplitude

D
Eight times the amplitude

Solution

A particle executing S.H.M completes a distance in one complete cycle in one time period which is equal to four times the amplitude.
Question 9
1 / -0

The value of phase at maximum displacement from the mean position of a particle in $$S.H.M.$$ is :

A
$$\pi /2$$

B
$$\pi$$

C
$$zero$$

D
$$2\pi$$

Solution
Question 10
1 / -0

The co-ordinates of a moving particle at a time $$t$$, are given by , $$x= 5\ sin\ 10t$$, $$y= 5\ cos\ 10t$$. The speed of the particle is:

A
$$25$$

B
$$50$$

C
$$10$$

D
None

Solution