Back to OSCILLATIONS

NEET UG PHYSICS MCQ: OSCILLATIONS

1 Easy

1 of 1

Application +4 | -1

A spring has a natural length of

50 \mathrm{~cm}

and a force constant of

2 \times 10^3 \mathrm{Nm}^{-1}

. A body of mass

10 \mathrm{~kg}

is suspended from it and the spring is stretched. If the body is pulled down to a length of

58 \mathrm{~cm}

and released, it executes simple harmonic motion. What is the net force on the body when it is at its lowermost position of its oscillation?

View answer and explanation

Correct option: C

60 \mathrm{~N}

Option C (

60 \mathrm{~N}

) is correct because the net force at the lowermost position is the restoring force of the spring minus the weight of the body.

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)

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.