Question

A string is stretched along the x-axis. There is a transverse disturbance along the string given by the equation $\varphi (x,t)=4sin\left(\frac{\pi x}{3}\right)cos\left(10\pi t\right)$where x is measured in centimeters and t in second. Both waves moving towards each other create this disturbance. Assume the velocities and amplitudes are equal for the two waves.

• A. Amplitude of each waves is 2 cm
• B. Velocity of each waves is 30 cm/sec
• C. Distance between two adjacent nodes is 3 cm
• D. Distance between two adjacent nodes is 2 cm

Answer 1

Answer: (A), (B), (C)

The correct options are
A

Amplitude of each waves is 2 cm

B

Velocity of each waves is 30 cm/sec

C

Distance between two adjacent nodes is 3 cm

If we use the trigonometric identity,

(i).... sin x+ sin y = 2 sin $\left(\frac{y+x}{2}\right)cos\left(\frac{y-x}{2}\right)$

The transverse perturbation can be decomposed into two transverse waves which move in different directions:

(ii) .....{ ${\varphi }_1(x,t)=Asin(kx-\omega t){\varphi }_2(x,t)=Asin(kx-\omega t)$

Since 2A = 4 , we immediately obtain A = 2 cm . The velocity of each wave is,

(iii) . . . . v=$\frac{\omega }{k}=\frac{10\pi }{\pi /3}$

The distance between two adjacent nodes can be derived from the condition sin(kx) = 0, which leads to:

(iv) . . . . kx = nx, n = 0, 1, 2 . . . . .

(v) or x = $\frac{x}{k}n=\frac{\lambda }{2}n$

The wavelength can be calculated by considering the given $\varphi (x,t)$ :

(vi) $\lambda =\frac{2\pi }{k}=\frac{2\pi }{\pi /3}=6$ cm

Therefore, the distance between two adjacent nodes (n = 1) is : l = $\frac{\lambda }{2}=3cm$