Question

What are the (a) amplitude, (b) frequency, and (c) phase constant of the oscillation shown in $FIGUREEX15.5?$

Short answer

a. amplitude$a=20cm$

b. frequencylocalid="1650197445764" $f=0.125Hz$

c. phase constantlocalid="1650285550846" role="math" ${150}^0$

Step-by-step solution

Given information

The given $FIGUREEX15.5$carries the information of amplitude, frequency and phase constant.

Explanation (part a)

The amplitude is the length of the peak from the time axis. From figure , it is clear that the amplitude is $a=20cm$

Explanation (part b)

The frequency is the number of waves per unit time. From $FIGUREEX15.5$, in $8s$the number of waves completed is localid="1650197099580" $1.$

Therefore, frequency is,

localid="1650197155998" $f=\frac{numberofwaves}{time}=\frac{1}{8s}=0.125Hz$

Explanation (Part c)

Suppose the equation of the wave is,

$x=a\sin \left(\omega t+\theta \right)$

Here, $\omega$is the angular frequency of the wave, and $\theta$is initial phase constant.

The time period of the wave is,

localid="1650197198931" $T=8s$

Therefore, the angular frequency is,

localid="1650197228672" $\omega =\frac{2\mathrm{\pi }}{T}=\frac{2\mathrm{\pi }}{8}=\frac{\mathrm{\pi }}{4}rad/s$

For initial phaselocalid="1650197283691" $t=0s$and$x=10cm$

The equation takes the form,

localid="1650285529821" $$\begin{gathered} 10cm=(20cm)\sin \left(\frac{\mathrm{\pi }}{4}\left(0s\right)+\theta \right) \\ \sin \left(\mathrm{\theta }\right)=\frac{1}{2} \\ \sin \theta =\frac{1}{2}=\sin {150}^0 \\ \mathrm{\theta }={150}^0 \end{gathered}$$

Therefore, the phase constant or the initial phase islocalid="1650285650108" ${150}^0$