First slide

Speed of wave

Question

A string of uniform cross-sectional area is hanging from one of tis ends as shown in figure. A transverse sinusoidal pulse is generated at the lower end and the pulse is propagating up the string. If the wavelength of wave at the mid-point C is $λ$ , what is the wavelength at the uppermost point A?

Difficult

A

$\frac{λ}{\sqrt{2}}$
B

$2 λ$
C

$\frac{λ}{2}$
D

$\sqrt{2} λ$

Solution

Velocity of wave at a distance x from the lower end is $V_{x} = \sqrt{gx}$ .
Frequency (f) of wave remains unchanged throughout the length of the string. Therefore wave length $λ_{x}$ at a distance x from the lower end is given by $λ_{x} = \frac{V_{x}}{f} = \frac{\sqrt{gx}}{f} \Rightarrow λ_{x} \propto \sqrt{x}$
$∴ \frac{λ_{A}}{λ_{C}} = \sqrt{\frac{x_{A}}{x_{C}}} \Rightarrow \frac{λ_{A}}{λ} = \sqrt{\frac{l}{l / 2}} \Rightarrow λ_{A} = \sqrt{2} λ$

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