Question

A mass is attached to the end of a spring and set into simple harmonic motion with an amplitude A on a horizontal frictionless surface. Determine the following in terms of only the variable A.
(a) Magnitude of the position (in terms of A) of the oscillating mass when its speed is 50% of its maximum value.
(b) Magnitude of the position (in terms of A) of the oscillating mass when the elastic potential energy of the spring is 50% of the total energy of the oscillating system.

Answer 1

Answer:

Explanation:

Assuming  motion of equation of SHM as

$x=A\cos \omega t$

where, $A=amplitude$

$\omega =angular\ frequency$

t=time

thus velocity at any Point x is given by

$v=\omega \Sqrt{A^2-x^2}$

when speed is 50 % of its maximum value

$v=0.5 A\omega$

$0.5 A\omega =\omega \Sqrt{A^2-x^2}$

$\frac{A^2}{4}=A^2-x^2$

$x=\frac{\sqrt{3}}{2}A$

(b)When Potential Energy is 50 % of total energy

Total Energy $=\frac{1}{2}kA^2$

potential energy at any x is given by $U=\frac{1}{2}kx^2$

$U=0.5\times \frac{1}{2}kA^2$

$\frac{1}{2}kx^2=0.5\times \frac{1}{2}kA^2$

$x=\frac{A}{\sqrt{2}}$