Answer to Question #161770 in Physics for RAJ

Question

A transverse traveling wave on a cord is represented by y(x,t) = 0.48 sin (4πx + 10πt) where y and x are in meters and t in seconds.

a) For this wave determine the wavelength, frequency, phase velocity (magnitude and direction), amplitude, maximum and minimum transverse speeds of particles of the cord.

b) What is the minimum length of the cord required for a standing wave to be formed when this wave meets its reflection?

c) Write the equation representing the standing wave.

What is the amplitude of vibration for a particle of the string located at x = 0.15 m?

Answer 1

(a)

$$\lambda=\dfrac{2\pi}{k}=\dfrac{2\pi}{4\pi\ m^{-1}}=0.5\ m.$$ $$f=\dfrac{\omega}{2\pi}=\dfrac{10\pi\ \dfrac{rad}{s}}{2\pi}=5\ Hz.$$ $$v=\dfrac{10\pi\ \dfrac{rad}{s}}{4\pi\ m^{-1}}=2.5\ \dfrac{m}{s},$$

The wave propagates along $x$-axis.

$$A=0.48\ m.$$

Let's take the derivative with respect to $t$:

$$v(x,t)=\dfrac{d}{dt}(y(x,t)),$$ $$v(x,t)=\dfrac{d}{dt}(0.48sin(4\pi x+10\pi t)=0.48\cdot10\pi cos(4\pi x+10\pi t),$$ $$v(x,t)=15.1 cos(4\pi x+10\pi t).$$

The maximum and minimum transverse speeds of particles of the cord will be when

$$cos(4\pi x+10\pi t)=\pm1.$$

Therefore,

$$v_{max}=15.1\ \dfrac{m}{s}, v_{min}=-15.1\ \dfrac{m}{s}.$$

(b) The minimum length of the cord required for a standing wave to be formed can be found from the formula:

$$L=\dfrac{1}{2}\lambda=\dfrac{1}{2}\cdot0.5\ m=0.25\ m.$$

(c) Let's consider two identical waves that move in opposite directions. The first wave has a wave function of $y_1(x,t)=0.48sin(4\pi x+10\pi t)$ and the second wave has a wave function $y_2(x,t)=0.48sin(4\pi x-10\pi t)$. The waves interfere and form a resultant wave:

$$y(x,t)=y_1(x,t)+y_2(x,t),$$ $$y(x,t)=0.48sin(4\pi x+10\pi t)+0.48sin(4\pi x-10\pi t).$$

Using the trigonometric identity

$$sin(\alpha \pm \beta)=sin\alpha cos\beta \pm cos\alpha sin\beta$$

we can write the equation representing the standing wave:

$$y(x,t)=2\cdot0.48sin(4\pi x)cos(10\pi t),$$ $$y(x,t)=0.96sin(4\pi x)cos(10\pi t).$$

Let's assume that initially at time $t=0$ two waves are in phase. Then, $cos(10\pi t)=1$ and we get the amplitude of vibration for a particle of the string located at x = 0.15 m:

$$y(x,t)=0.96sin(4\pi\cdot0.15\ m)=0.031\ m.$$