A mass of 1.53 kg is attached to a spring and the system is undergoing simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. What is the speed of the mass when it is 3.00 cm from its equilibrium position?
A) 0.0368 m/s
B) 0.551 m/s
C) 0.421 m/s
D) 0.842 m/s
E) 0.919 m/s

Question

Quizplus · Accepted Answer

We can use the equation for the speed of an object undergoing simple harmonic motion at a given displacement from the equilibrium position:

v = ω√(A^2 - x^2)

where ω is the angular frequency (ω = 2πf) and x is the displacement from the equilibrium position.

Plugging in the given values, we get:

ω = 2πf = 2π(1.95 Hz) ≈ 12.24 rad/s
A = 7.50 cm = 0.075 m
x = 3.00 cm = 0.030 m

Thus, the speed of the mass at a displacement of 3.00 cm is:

v = ω√(A^2 - x^2) ≈ 0.842 m/s

Therefore, the correct answer is D.

A Mass of 1