Question

A particel is performing simple harmonic motion it's maximum velocity is 3m/s and amplitude of motion is 0.2 m. calculate the Time period of motion​

Answer 1

Answer:

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An object moving along the x-axis is said to exhibit simple harmonic motion if its position as a function of time varies as

x(t) = x0 + A cos(ωt + φ).

The object oscillates about the equilibrium position x0. If we choose the origin of our coordinate system such that x0 = 0, then the displacement x from the equilibrium position as a function of time is given by

x(t) = A cos(ωt + φ).

A is the amplitude of the oscillation, i.e. the maximum displacement of the object from equilibrium, either in the positive or negative x-direction. Simple harmonic motion is repetitive. The period T is the time it takes the object to complete one oscillation and return to the starting position. The angular frequency ω is given by ω = 2π/T. The angular frequency is measured in radians per second. The inverse of the period is the frequency f = 1/T. The frequency f = 1/T = ω/2π of the motion gives the number of complete oscillations per unit time. It is measured in units of Hertz, (1 Hz = 1/s).

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Answer 2

vmax = 3m/s

and amplitude = 0.2 m

vmax = a w

so 3/0.2 = w

w = 15

now we know that .

w= 2π/t

so t= 2π /w

t = 6.3/15

t =~o.4 sec