Question

a mass oscillates on a spring with a period of 0.73 s and an amplitude of 5.4 cm. write an equation giving x as a function of time, assuming the mass starts at x = a at time t = 0.

a. x=(2.4rad/s)cos(1.1cm)
b. x=(0.73s)cos9.3cm
c. x=(1.8cm)cos(1.8rad/s)
d. x=(5.4cm)cos(8.6rad/s)
e. x=(10.7rad/s)cos(5.1rad/s)

Explanation:

Step1: Recall the general equation for simple - harmonic motion

The general equation for the position of an object in simple - harmonic motion is $x = A\cos(\omega t+\varphi)$, where $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is the time, and $\varphi$ is the phase constant. Given that the mass starts at $x = A$ at $t = 0$, $\varphi=0$.

Step2: Calculate the angular frequency $\omega$

The relationship between the period $T$ and the angular frequency $\omega$ is $\omega=\frac{2\pi}{T}$. Given $T = 0.73\ s$, then $\omega=\frac{2\pi}{0.73}\ s^{-1}\approx8.6\ rad/s$. The amplitude $A = 5.4\ cm$. So the equation for $x$ as a function of time is $x=(5.4\ cm)\cos(8.6\ rad/s\times t)$.

Answer:

D. $x=(5.4\ cm)\cos(8.6\ rad/s\times t)$