A mass on a spring vibrates in simple harmonic motion at a frequency of 4.26 Hz and an amplitude of 3.00 cm.If a timer is started when its displacement is a maximum (hence x = 3.00 cm when t = 0),what is the acceleration magnitude when t = 3.74 s? A)zero B)0.732 m/s² C)39.2 m/s² D)19.6 m/s² E)10.5 m/s²

Question

Quizplus · Accepted Answer

The acceleration of the mass during simple harmonic motion is given by the equation a = -ω^2x, where ω is the angular frequency and x is the displacement from equilibrium. 
First, we need to find the angular frequency, ω:

ω = 2πf = 2π(4.26 Hz) = 26.8 rad/s

Next, we can use the given displacement and time to find the current position of the mass:

x = A cos(ωt) = (3.00 cm) cos(26.8 rad/s * 3.74 s) = -1.81 cm

Note that the negative sign indicates that the mass is currently moving in the negative direction.

Now we can find the acceleration:

a = -ω^2x = -(26.8 rad/s)^2(-1.81 cm) = 19.6 m/s^2

Therefore, the acceleration at t = 3.74 s is 19.6 m/s^2, which corresponds to choice D.

A Mass on a Spring Vibrates in Simple Harmonic Motion