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Firlakuza [10]
3 years ago
8

A mass is attached to the end of a spring and set into simple harmonic motion with an amplitude A on a horizontal frictionless s

urface. 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.
Physics
1 answer:
IgorLugansk [536]3 years ago
5 0

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}}

   

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