Home
Class 11
PHYSICS
A particle executing SHM crossed points ...

A particle executing SHM crossed points A and B with the same velocity. Having taken 3 s in passing from A to B, it returns to B after another 3s. The time period is :

A

15s

B

6s

C

12s

D

9s

Text Solution

Verified by Experts

The correct Answer is:
C
Time period of Oscillation = `2xx` (time taken to go from A to B + the next time taken to return at B)
`=2xx(3+3)`
`=2xx6`
Time period = 12s
Promotional Banner

Topper's Solved these Questions

  • OSCILLATIONS

    FULL MARKS|Exercise TEXTUAL EVALUATION SOLVED ( II SHORT ANSWER QUESTION )|15 Videos

  • OSCILLATIONS

    FULL MARKS|Exercise TEXTUAL EVALUATION SOLVED ( III LONG ANSWER QUESTIONS )|10 Videos

  • OSCILLATIONS

    FULL MARKS|Exercise ADDITIONAL QUESTIONS SOLVED (IV NUMERICAL PROBLEMS )|10 Videos

  • NATURE OF PHYSICAL WORLD AND MEASUREMENT

    FULL MARKS|Exercise ADDITIONAL QUESTIONS SOLVED ( SHORT ANSWER QUESTIONS (2 MARK))|20 Videos

  • PROPERTIES OF MATTER

    FULL MARKS|Exercise Additional Questions Solved - Numerical Questions|19 Videos

Similar Questions

Explore conceptually related problems

A particle executing SHM crossed points A and B with the same velocity. Having taken 2 s in passing from A to B, it returns to B after another 3s. The time period is :

A particle is executing SHM along a straight line. Its velocities at dsitances x_1 and x_2 from the mean position are V_1 and V_2 , respectively. Its time period is

A particle executing SHM has an acceleration of 64cm//s^(2) with its displacement is 4 cm. Its time period, in seconds is……………

The position velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2cm, 1ms^-1 and 10ms^-2 at a certain instant. Find the amplitude and the time period of the motion.

A particle starts from a point A and travels along the solid curve shown in figure. Find approximately the position B of the particle such that the average velocity between the positions A and B has the same direction as the instantaneous velocity at B.

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then find its time period in seconds.

The shortest distance travelled by a particle executing SHM form mean position in 2 seconds is equal to ( sqrt(3))/(2 ) times of its amplitude . Determine its time period .