NEET UG PHYSICS MCQ: OSCILLATIONS

1 Easy

1 of 1

Application +4 | -1

The motion of particle executing S.H.M is given by

x = 0.01\,\,\sin \pi (100\,t + 0.05),

where x in metres and time is in seconds. The time period is

View answer and explanation

Correct option: B

0.02 s

Option B (0.02 s) is correct because the time period is calculated from the angular frequency, which is the coefficient of time in the sine function's argument.

More Questions from OSCILLATIONS

Practice All Chapter MCQs →

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)

and

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.