Question

The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by

$y=\left(0.4cm\right)sin\left[\right(0.314c{m}^{-1}\left)x\right]cos\left[\right(600\pi s^{-1}\left)t\right].$

(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 waavelength and the speed of two travelling waves that can interfere to give this vibration ?

Answer 1

The stationary wave equationis given by

$y=\left(0.4cm\right)sin\left[\right(0.314!c{m}^{-1}\left)x\right]cos\left[\right(6.00\pi s^{-1}\left)t\right]$

(a) $\omega =600\pi$

$\Rightarrow 2\pi f=600\pi$

$\Rightarrow f=300Hz$

Wavelength,

$\lambda =\frac{2\pi }{0.314}=\frac{(2\times 3.14)}{0.314}$

$=20cm$

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

(c) Length of the string

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

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

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

Since, $\lambda$ and v are the wavelength and velocity of the waves that interfere to give this vibration.

$\lambda =20cm$

$v=\frac{\omega }{k}=\frac{600\pi }{\left(\frac{\pi }{10}\right)}$

$=6000cm/sec=60m/s$