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 40% of its maximum value. A (b) Magnitude of the position (in terms of A) of the oscillating mass when the elastic potential energy of the spring is 40% of the total energy of the oscillating system. А

Answer 1

Part a.

Given : speed of oscillating mass = 40% of maximum speed.

Speed can be calculated as V = w √ ( A² - x² ) .... (1)

Where, w is the angular frequency or angular velocity

A is the amplitude

x is the position of mass.

Now according to above equation when x= 0 , speed will be maximum

Hence , v(max) = w. A

Now given speed is 40% of maximum

V = (40/100) wA = 0.4 wA

To find x, substitute V in eq (1)

0.4 wA = w √ ( A² - x² )

Solving this we get,

x = 0.916 A ( Answer)

Part b.

Elastic potential energy of system V = 40% of total energy

Total energy = maximum potential energy = (1/2) kA²

Hence, V = (40/100) Vmax = 0.4 (1/2) kA² = 0.2 kA²

also, potential energy at any point X is given by

V= (1/2) kX²

Therefore,

0.2 kA² = (1/2) kX²

Solving for X,

X = 0.2 A ( Answer)

I hope it helps you. For further queries please ask in comments.