Question

The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by
y=(0·4 cm) sin[(0·314 cm-1) x] cos [(600π s-1) t].
(a) What is the frequency of vibration? (b) What are the positions of the nodes? (c) What is the length of the string? (d) What is the wavelength and the speed of two travelling waves that can interfere to give this vibration?

Answer 1

Frequency of Vibration is 300 Hz

Explanation:

The stationary wave equation is given by -

y = $(0.4\ cm)\ sin [(0.314 cm - 1)x]\cos [(600 \pi s^-^1)t]$

a.) $\omega = 600 \pi$

$2\pi f = 600\pi$

f = 300 Hz

wavelength, \lambda = $2\pi f = \frac {(2 \times 3.14)}{0.314}$ = 20 cm

(b) Therefore nodes are located at, 0, 10 cm, 20 cm, 30 cm

(c) Length of the string = $\frac {3\pi}{\lambda}$

= $3 \times \frac {20}{2}$= 30 cm

(d) y = $0.4\ sin (0.314x)\ cos(600 \pi t)$

$0.4\ sin {(\frac {\pi}{10})x}\cos (600\pi t)$

since, $\lambda$ and v are the wavelength and velocity of the waves that interfere to give this vibration $\lambda$ = 20cm

v = $\frac {\omega}{k} = 6000\ cmsec^-^1$

= 60 $ms^-^1$