Question

A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibration is 1⋅0 and the displacement becomes zero 200 times per second. The linear mass density of the string is 0⋅10 kg m−1 and it is kept under a tension of 90 N. (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation. (c) Find the velocity and acceleration of the particle at x = 50 cm at time t = 10 ms.

Answer 1

Wave Speed is 30 m/s

Explanation:

Amplitude, A = 1 cm,  

Tension T = 90 N  

Frequency, f = $\frac {200}{2}$ = 100 Hz  

Mass per unit length, m = $0.1 \frac {kg}{mt}$

a) ⇒ V = $\sqrt{T}{m} = 30 ms^-^1 \lambda$

=  $\frac {v}{f}$

= $\frac {30}{100}$

= 0.3 m = 30 cm  

b) The wave equation y = (1 cm) $cos 2\pi$ ( $\frac {t}{0.01}$ s) – ( $\frac {x}{30} cm$) [because at x = 0, displacement is maximum]  

c) y = $1\ cos 2\pi(\frac {x}{30} - \frac {t}{0.01})$

⇒ v = $\frac {dy}{dt} \frac {dy}{dt}$

= $( \frac {1}{0.01})2\pi sin 2\pi {( \frac {x}{30}) - ( \frac {t}{0.01})} a$  

= $\frac {dv}{dt}$

= $-\ {4\pi2}{(0.01)2} cos 2\pi { \frac {x}{30}(\frac {x}{30}) - \frac {t}{0.01}}$

When, x = 50 cm, t = 10 ms = 10 $\times 10 - 3$ sx  

= $\frac {2\pi}{0.01} sin 2\pi {(\frac {5}{3} - (\frac {0.01}{0.01})}$

= $\frac {p}{0.01} sin (2\pi \times \frac {2}{3})$

= $(\frac {1}{0.01}) sin \frac{4\pi}{3}$

= $- 200 \pi sin \frac {\pi}{3}$  

= $- 200 \pi \times \frac {\sqrt3}{\sqrt2}$

= $544\ cms^-^1$

= $5.4\ ms^-^1$

Similarly a = ${\frac {4\pi2}{(0.01)2} cos 2\pi {(\frac {5}{3}) - 1}$

= $4\pi2 \times 104 \times \frac {1}{2}$

⇒ $2 \times 105\ cms^-^2$

= $2\ kms^-^2$