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NEET UG PHYSICS MCQ: OSCILLATIONS

1 Easy

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Application +4 | -1

A particle starts oscillation from origin on

x

-axis with period of oscillation (6) sec and amplitude

A

. If time taken by particle to reach from

x=A

to

x=\frac{\sqrt{3}}{2} A

for the first time is

\tau

then. Value of

6 \tau

is sec.

View answer and explanation

Correct option: C

3

Option C (3) is correct because the time taken to go from A to

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

is T/12, and T = 6s, so

\tau

= 6/12 = 0.5s. Therefore, 6

\tau

= 3.

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A mass m is suspended from a spring of length l and force constant K. The frequency of vibration of the mass is

{f_1}

. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is

{f_2}

. Which of the following relations between the frequencies is correct The time period of a loaded spring on earth is 6s. On the surface of moon, the time period of the same loaded spring will be. Under the action of a force

F=-k x^3

, the motion of a particle is

(k=

a positive constant

)

S-I: A particle executes SHM with frequency

f

. The frequency with which total mechanical energy of particle oscillates is

2 f

.
S-II: Both kinetic as well as potential energy of particle in SHM oscillate with a frequency

2 f.

The phase difference (in radian) between two SHM whose equations are

y_1=10 \sin \left(10 \pi t+\frac{\pi}{3}\right)

y_2=12 \sin \left(8 \pi t+\frac{\pi}{4}\right)

t=0.5 \mathrm{~s}

is: A vertical spring block system is made to oscillate.
Statement-1: Its time period on earth is more than that on the moon.
Statement-2: Its extension on earth (in equilibrium) is more than that on the moon.