Home
Class 12
PHYSICS
A particle starts from mean position and...

A particle starts from mean position and moves towards positive extreme as show below. Find the equation of the `SHM`, Amplitude of `SHM` is `A`.

Text Solution

Verified by Experts

General equation of `SHM` can be written as `x = Asin(omegat+phi)`
At `t = 0, x = 0 , :. 0 = Asinphi, :. phi = 0, pi , phiin[0,2pi]`
Also, at `t = 0, v = +ve , :. Aomegacosphi = "maximum" +ve` or `phi = 0`
Hence, if the particle is at mean position at `t = 0` and is moving towards `+ve` extreme, then the equation of `SHM` is given by `x = Asinomegat`

Similarly, for particle moving towards `-ve` extreme then `phi = pi`
`:.` equation of `SHM` is `x = Asin(omegat + pi)` or, `x = -Asinomegat`
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    ALLEN |Exercise SOME WORKED OUT EXAMPLES|29 Videos

  • SIMPLE HARMONIC MOTION

    ALLEN |Exercise Exercise-01|117 Videos

  • RACE

    ALLEN |Exercise Basic Maths (Wave Motion & Dopplers Effect) (Stationary waves & doppler effect, beats)|25 Videos

  • TEST PAPER

    ALLEN |Exercise PHYSICS|4 Videos

Similar Questions

Explore conceptually related problems

A particle starts from point x = (-sqrt(3))/(2)A and move towards negative extreme as shown, (a) Find the equation of the SHM (b) Find the time taken by the particle to go directly from its initial position to negative extreme. (c) Find the time taken by the particle to reach at mean position.

A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second it travels a distance b in the same direction. The amplitude of the SHM is

Amplitude of an SHO is A. When it is at a distance y from the mean position of the path of its oscillation, the SHO receives blow in the direction of its motion, which doubles its velocity instantaneously. Find the new amplitude of its oscillations.

In SHM………….quantities are always positive.

Write the equation of SHM for the sitution shown below :

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 its time period in second is…….

A particle executes simple harmonic with an amplitude of 5 cm. When the particle is at 4 cm from the mean position the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is

A point particle if mass 0.1 kg is executing SHM of amplitude 0.1 m . When the particle passes through the mean position, its kinetic energy is 8 xx 10^(-3)J . Write down the equation of motion of this particle when the initial phase of oscillation is 45^(@) .

A point particle if mass 0.1 kg is executing SHM of amplitude 0.1 m . When the particle passes through the mean position, its kinetic energy is 8 xx 10^(-3)J . Write down the equation of motion of this particle when the initial phase of oscillation is 45^(@) .

A particle simple harmonic motion completes 1200 oscillations per minute and passes through the mean position with a velocity 3.14 ms^(-1) . Determine displacement of the particle from its mean position. Also obtain the displacement equation of the particle if its displacement be zero at the instant t = 0 .