Question

A mass is attached to an ideal spring. At time t = 0 the spring is at its natural length and the mass is given an initial velocity; the period of the ensuing (one-dimensional) simple harmonic motion is T . At what time is the power delivered to the mass by the spring first a maximum?

Answer 1

t = T/4

Step-by-step explanation:

The power delivered to the mass by the spring is work done by the spring per second.

`P = (dW)/(dt)`

The work done by the spring is equal to the elastic potential energy stored in the spring.

`U = (1)/(2)kx^2`

The maximum energy stored in the spring is at the amplitude of the oscillation.

`U_(max) =(1)/(2)kA^2`

So the first time the mass reaches to its amplitude can be found by the following equation of motion:

`x = A\cos(\omega t + \phi)\\\phi = \pi/2 ~because ~at ~t= 0, ~ x = 0\\0 = A\cos(0 + \pi/2)\\x = A\cos(\omega t + \pi/2)`

When the mass reaches the amplitude:

`A = A\cos(\omega t + \pi/2)\\1 = \cos(\omega t + \pi/2)\\\omega t + \pi/2 = \pi`

because cos(π) = 1.

`\omega t = \pi/2`

Using ω = 2π/T,

`\omega t = \pi/2\\(2\pi)/(T)t = \pi/2\\t = (T)/(4)`