Question

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

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

i. What is the frequency of vibration?

ii. What are the positions of the nodes?

iii. What is the length of the string?

iv. What is the wavelength and the speed of two travelling waves that can interfere to give this vibration?

• A. $300Hz,0,10cm,20cm,30cm,L=30cm,\lambda =20cm,v=60m{s}^{-1}$
• B. $600Hz,0,15cm,30cm,L=30cm,\lambda =20cm,v=60m{s}^{-1}$
• C. $300Hz,0,10cm,20cm,L=20cm,\lambda =20cm,v=60m{s}^{-1}$
• D. None of these

Answer 1

The correct option is A

$300Hz,0,10cm,20cm,30cm,L=30cm,\lambda =20cm,v=60m{s}^{-1}$

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

From the equation $\omega =600\pi$

$f=\frac{\omega }{2\pi }=300Hz$

$k=0.314$

$\lambda =\frac{2\pi }{k}=\frac{2\times 3.14}{0.314}=20cm;v=f\lambda =6000cm{s}^{-1}$ as it is the third harmonic

$f=f_3=\frac{3v}{2L}$

$300=\frac{3\times 6000}{2\times L}$

$L=30cm$

If we draw this we can see the position of nodes

Nodes occur at the ends and 10 cm and 20 cm. two identical waves travelling in opposite direction would create such standing wave their speed and wavelengths would be 6000

$cm{s}^{-1}=60m{s}^{-1}and20$

$cm.as,y_1=0.2sin(0.314\times -600\pi t)$

$y_2=0.2sin(0.314\times +600\pi t+\pi )$

would add up to given equation.