A transverse sinusoidal wave of amplitude a , wavelength ( $λ$ ) and frequency $f$ is travelling on the stretched string. The maximum speed of any point on the string is $v / 10$ where v is the speed of propagation of the wave . If a $= 10^{- 3} m a n d v = 10 m / s,$ then $λ$ and $f$ are given by the speed of

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$f = \frac{10^{3}}{2 π} Hz$

$λ = 10^{- 3} m$

$f = 10^{4} Hz$

$λ = 2 π \times 10^{- 2} m$

answer is A, C.

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Detailed Solution

Maximum speed of any point on the string $= a ω$

$∴ \frac{v}{10} = \frac{10}{10} = 1 (G i v e n, v = 10 m / s) ∴ 2 π a f = 1 \Rightarrow f = \frac{1}{2 π a} (a = 10^{- 3} m, given)$

$T h e r e f o r e f = \frac{1}{2 π \times 10^{- 3}} = \frac{10^{3}}{2 π} Hz S p e e d o f w a v e v = f λ ∴ (10 m / s) = (\frac{10^{3}}{2 π}) λ ∴ λ = 2 π \times 10^{- 2} m$

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