Question

A wave pulse having wavelength do is generated at
| lower end of a vertical string (Mass M, length L).
Wavelength of pulse when it reaches at top will be​

Answer 1

Answer:

Wavelength of the wave in the string at the top end is given as

$\lambda = \sqrt{\frac{m + M}{m}}\lambda_o$

Explanation:

As we know that the frequency of the wave is always constant during its propagation into the medium

so here the speed of wave at lowest point of the string is given as

$v = \sqrt{\frac{T}{\mu}}$

where we have

$T = mg$

now at the top end of the string we have

$T =(m + M)g$

so we have

$\frac{v_1}{\lambda_1} = \frac{v_2}{\lambda_2}$

$\sqrt{\frac{mg}{\mu}}\frac{1}{\lambda_o} = \sqrt{\frac{(m+M)g}{\mu}}\frac{1}{\lambda}$

so we have

$\lambda = \sqrt{\frac{m + M}{m}}\lambda_o$

#Learn

Topic : String Wave

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