Question

An object attached to a spring vibrates with simple har- monic motion as described by Figure P15.64. For this motion, find (a) the amplitude, (b) the period, (c) the * (cm) 2.00 1.00 0.00 21(s) 3 4 5 6 -1.00 -2.00 angular frequency, (d) the maximum speed, (e) the maximum acceleration, and (f) an equation for its posi- tion x as a function of time.

Answer 1

-T x (cm) 2.00 1.00 t (s) 6 0.00 -1.00 -2.00

(a)

One complete wave is highlighted in the above figure. Note that the maximum value of $x$ is given by $A=2cm=0.02m$.

(b)

One complete oscillation is completed in time $T=4s$

(c)

The angular frequency is given by

$\omega=\frac{2\pi}{T}$

$\omega=\frac{2\pi}{4\ s}=\frac {\pi}{2}\ s^{-1}$

$\omega=1.57\ s^{-1}$

(d)

In general, a simple harmonic motion is given by

$x=Asin(\omega t+\phi)$

To find velocity, differentiate it once with respect to time

$v=\dot{x}=\omega Acos(\omega t+\phi)$

The maximum speed in a simple harmonic oscillation is given by

$v_{max}=\omega A$

$v_{max}=1.57\ s^{-1} \times 0.02m=0.0314\ ms^{-1}$

(e)

To find acceleration in simple harmonic motion, differentiate velocity with respect to time

$a=\dot{v}=-\omega ^2Asin(\omega t+\phi)$

The maximum acceleration in a simple harmonic oscillation is given by

$a_{max}=\omega ^2A$

$a_{max}=(1.57\ s^{-1})^2\times 0.02m=6.28\times 10^{-4}\ ms^{-2}$

(f)

A simple harmonic motion is given by

$x=A sin (\omega t+\phi)$

In the figure, note that $x=0$ at $t=0$

$0=A sin (\phi)$

Since A=0.02m we conclude  $sin (\phi)=0$

$\Rightarrow \phi=0$

The given simple harmonic motion is

$x=A\ sin(\omega t)=2.00\ sin(1.57\ t)\ cm$