Question

You and a friend launch waves at opposite ends of the stretched string after class. You shake the string at 5 Hz with a 15 cm amplitude, and your acquaintance shakes the other end at 2 Hz with a 10 cm amplitude. What is the largest (most positive) displacement in cm of the string at any point at any time? Explain. b. Is your wave faster? Explain.

Answer 1

Let one wave be

$y_1=A_1\cos(k_1x-\omega_1t)$

and the other be

$y_2=A_2\cos\left[k_2(L-x)-\omega_2t\right]=A_2\cos(k_2x+\omega_2t-\delta)$

where $\delta=k_2L$ and L is the length of string. Also, k's and $\omega's$ are the angular wavenumber and angular frequency respectively. Superposition gives the net displacement as as function of position and time.

$y=y_1+y_2=A_1\cos(k_1x-\omega_1t)+A_2\cos(k_2x+\omega_2t-\delta)$

Now we just need to maximise this.

The maximum is $A_1+A_2=25\textup{ cm}$

when

$k_1x-\omega_1t=2n\pi,\quad k_2x+\omega_2t-\delta=2m\pi$

for integers n and m. Express k's in terms of speed of the wave v

$\omega_1\left (\frac{x}{v}-t \right )=2n\pi,\quad \omega_2\left (\frac{x}{v}+t \right )-\delta=2m\pi$

$\frac{x}{v}-t=\frac{n}{f_1},\quad \frac{x}{v}+t-\frac{L}{v}=\frac{m}{f_2}$

Solve the 2 equations simultaneously, to get

$x=\frac{v}{2}\left (\frac{L}{v}+\frac{n}{f_1}+\frac{m}{f_2} \right ),\quad t=\frac{1}{2}\left (\frac{L}{v}-\frac{n}{f_1}+\frac{m}{f_2} \right )$

Exactly for what value of x and t will it occur will depend on the values of speed of the wave and the phase $\delta.$

(b) Both waves have the same speed because it is the property of string.

$v=\sqrt{\frac{T}{\mu}}$