An impulsive force gives an intial velocity of `1.0 ms^(-1)` to the mass m in the unstretched spring position. What is the amplitude of motion ? Given that x is a function of time to for the oscillating mass. Give `m= 3kg and k=1200 Nm^(-1)`.

Explanation

Here initial velocity in unstretched position.
`v=-1.0 ms^(-1)`
Clearly, `v_("max")=1.0 ms^(-1)`
Given `k=1200 Nm^(-1), m=3 kg`
Also `omega =sqrt((k)/(m))= sqrt((1200)/(3))= "rad" s^(-1)`
Amplitude `A=(v_("max"))/(omega)=(1.0)/(20)=(1)/(20)m=5 cm`
As the motion starts from the understand position, the expression for the displacement can be written as,
`x=A sin omega t=5 sin 20t`
As initial impulse is naegative, the displacement is towards negative X-axis, so,
`x=-5 sin 20t`