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

Answer:

t = T/4

Explanation:

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

$P = \frac{dW}{dt}$

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

$U = \frac{1}{2}kx^2$

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

$U_{max} =\frac{1}{2}kA^2$

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,

$\omega t = \pi/2\\\frac{2\pi}{T}t = \pi/2\\t = \frac{T}{4}$