Home
Class 11
PHYSICS
A particle executes S.H.M. between x = -...

A particle executes S.H.M. between `x = -A` and `x = + A`. The time taken for it to go from 0 to A/2 is `T_(1)` and to go from A/2 to A is `T_(2)`. Then

A

`T_(1) lt T_(2)`

B

`T_(1) gt T_(2)`

C

`T_(1)=T_(2)`

D

`T_(1)=2T_(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Any SHM is given by the equation `x=sinomegat`, where x is the displacement of the body any instant t. A is the amplitude and `omega` is the angular frequency.
when `x=0,omegat_(1)=0`
`thereforet_(1)=0`
when `x=A//2,omegat_(2)=pi//6,t_(2)=pi//6omega`
when `x=A,omegat_(3)=pi//2,t_(3)=pi//2omega`
Time taken from o to A/2 will be
`t_(2)-t_(1)=(pi)/(6omega)=T_(1)`
time taken from A/2 to A will be
`t_(3)-t_(2)=(pi)/(2omega)-(pi)/(6omega)=(2pi)/(6omega)=(pi)/(3omega)=T_(2)`
Hence `T_(2)gtT_(1)`
Promotional Banner

Topper's Solved these Questions

  • OSCILLATIONS

    NCERT FINGERTIPS ENGLISH|Exercise Exemplar Problems|9 Videos

  • OSCILLATIONS

    NCERT FINGERTIPS ENGLISH|Exercise Assertion & Reason|15 Videos

  • OSCILLATIONS

    NCERT FINGERTIPS ENGLISH|Exercise Assertion And Reason|15 Videos

  • MOTION IN A STRAIGHT LINE

    NCERT FINGERTIPS ENGLISH|Exercise NCERT Exemplar|6 Videos

  • PHYSICAL WORLD

    NCERT FINGERTIPS ENGLISH|Exercise Assertion And Reason|10 Videos

Similar Questions

Explore conceptually related problems

A particle executing SHM of amplitude 4 cm and T=4 s .The time taken by it to move from positive extreme position to half the amplitude is

A particle undergoes S.H.M. having time period T . The time taken in 3 /8th oscillation is

A particle executes SHM in accordance with x=A"sin"omegat . If t_(1) is the time taken by it to reach from x=0 to x=sqrt(3)(A//2) and t_(2) is the time taken by it to reach from x=sqrt(3)//2 A to x=A , the value of t_(1)//t_(2) is

A particle performs SHM about x=0 such that at t=0 it is at x=0 and moving towards positive extreme. The time taken by it to go from x=0 to x=(A)/(2) is____ time the time taken to go from x=(A)/(2) to A . The most suitable option for the blank space is

A particle executes S.H.M with time period 12 s. The time taken by the particle to go directly from its mean position to half its amplitude.

A particle executes SHM with a time period of 4 s . Find the time taken by the particle to go from its mean position to half of its amplitude . Assume motion of particle to start from mean position.

Find the time taken by the particle in going from x = (A)/2) to x=A Hint : Time taken to go from x=0 to x =A Time taken to go from x=0 to x= (A)/(2) Required time = t_(1) - t_(2)

A particle is performing SHM of amplitude 'A' and time period 'T'. Find the time taken by the particle to go from 0 to A//2 .

A pariticle is performing SHM of amplitude "A" and time period "T" . Find the time taken by the period to go from 0 to A//2 .

A pariticle is performing SHM of amplitude "A" and time period "T" . Find the time taken by the period to go from 0 to A//root 2 .