Question

What are the (a) amplitude, (b) frequency, and (c) phase constant of the oscillation shown in $FIGUREEX15.6?$

Short answer

a. amplitude $a=10cm$

b. frequency $f=0.25Hz$

c. phase constant${120}^0$

Step-by-step solution

Given information

From the given $FIGUREEX15.6$amplitude, frequency and phase constant.

Explanation (part a)

The amplitude is the length of the peak from the time axis. From , it is clear that the amplitude is$a=10cm$

Explanation (part b)

The frequency is simply the number of waves per unit time. From $FIGUREEX15.6$, in $8s$the number of waves completed is$2.$

Therefore, frequency is,

localid="1650282912538" $f=\frac{numberofwaves}{time}=\frac{1}{4s}=0.25Hz$

Explanation (Part c)

Suppose the equation of the wave is,

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

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

The time period of the wave is,

$T=4s$

Therefore, the angular frequency is,

$\omega =\frac{2\mathrm{\pi }}{T}=\frac{2\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{2}rad/s$

Let's take what happens atlocalid="1650282979393" $t=0s$

The equation takes the form,

localid="1650283019843" $$\begin{gathered} -5cm=(10cm)\cos \left(\frac{\mathrm{\pi }}{2}\left(0s\right)+\theta \right) \\ \cos \left(\mathrm{\theta }\right)=-\frac{1}{2} \\ \mathrm{cos\theta }=-\frac{1}{2}=\cos {120}^0 \\ \mathrm{\theta }={120}^0 \end{gathered}$$

Therefore, the initial phase constant or the initial phase is${120}^0$