Home
Class 12
PHYSICS
Find the time taken by the paritcle in g...

Find the time taken by the paritcle in going from `x=0` to `x= (A)/(2)` where A is the amplitude.

Text Solution

AI Generated Solution

To find the time taken by a particle in simple harmonic motion (SHM) to go from \( x = 0 \) to \( x = \frac{A}{2} \), we can follow these steps: ### Step 1: Write the equation of motion for SHM The displacement \( x \) of a particle in SHM can be expressed as: \[ x = A \sin(\omega t) \] where: ...
Promotional Banner

Topper's Solved these Questions

  • OSCILLATIONS

    AAKASH INSTITUTE|Exercise Try Yourself|86 Videos

  • OSCILLATIONS

    AAKASH INSTITUTE|Exercise ASSIGNMENT ( SECTION -A)|58 Videos

  • NUCLEI

    AAKASH INSTITUTE|Exercise ASSIGNMENT (SECTION-D)|10 Videos

  • PHYSICAL WORLD

    AAKASH INSTITUTE|Exercise ASSIGNMENT (Section-B)|5 Videos

Similar Questions

Explore conceptually related problems

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

A particle executes a simple harmonic motion of time period T. Find the time taken by the particle to go directly from its mean position to half the 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.

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

A particle executes SHM with a time period of 12s. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

A particle of mass 2 kg executing SHM has amplitude 10 cm and time period 1s. Find (a) the angular frequency (b) the maximum speed (c ) the maxmum acceleration (d) the maximum restoring force ( e) the speed when the displacement from the mean position is 8 cm (f) the speed after (1)/(12)s the particle was at the extreme position (g) the time taken by the particle to go directly from its mean position to half the amplitude (h) the time taken by the particle to go directly from its extreme position to half the amplitude.

Assertion : Time taken by a particle in SHM to move from x = A to x = sqrt(3A)/(2) is same as the time taken by the particle to move from x = sqrt(3A)/(2) to x = (A)/(2) . Reason : Corresponding angles rotated is the reference circle are same in the given time intervals.

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.

The acceleration of a particle moving along x -axis is a= -100x .It is released from rest at x=2. Here a and x are in S.I.units.The minimum time taken by particle to go from x=2 to x=0 is (pi)/(alpha) seconds.Find the value of (alpha)/(4) .