A mass on a spring vibrates in simple harmonic motion at a frequency of 3.2 Hz and an amplitude of 6.4 cm.If a timer is started when its displacement is a maximum (hence x = 6.4 cm when t = 0),what is the displacement of the mass when t = 3.3 s?
A)zero
B)-5.95 cm
C)-2.36 cm
D)2.53 cm

Question

Quizplus · Accepted Answer

The displacement $x$ of a mass in simple harmonic motion can be described by the equation $x = A \cos(2\pi ft + \phi)$, where $A$ is the amplitude, $f$ is the frequency, $t$ is the time, and $\phi$ is the phase constant. Given that the timer starts when the displacement is at a maximum ($x = 6.4$ cm when $t = 0$), the phase constant $\phi = 0$. Thus, the equation simplifies to $x = A \cos(2\pi ft)$. Substituting the given values $A = 6.4$ cm, $f = 3.2$ Hz, and $t = 3.3$ s, we get $x = 6.4 \cos(2\pi \times 3.2 \times 3.3)$. Calculating this gives $x \approx -5.95$ cm.

A Mass on a Spring Vibrates in Simple Harmonic Motion