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Standard XII

Physics

Wave in Negative X-Direction

A string of m...

Question

A string of mass $0.2 k g / m$ and length $l = 0.6 m$ is fixed at both ends and stretched such that it has a tension of $80 N$ . The string vibrates in $3$ segments with maximum amplitude of $0.5 c m$ . The maximum transverse velocity amplitude is

1.57 m / s

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6.28 m / s

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3.14 m / s

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9.42 m / s

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Solution

The correct option is C $1.57 m / s$
Frequency of the wave traveling through the stretched wire with tension $T$ and mass per unit length $m$ is given by(string vibrates in $3$ segments):
$n = 3 / 2 L (\sqrt{T / m}) = 50 H z$ .
So, the maximum transverse velocity amplitude is:
$V_{m a x} = a_{m a x} ω = 2 π a_{m a x} n = 1.57 m / s$ .

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A string of mass 0.2kg/m and length l=0.6m is fixed at both ends and stretched such that it has a tension of 80N. The string vibrates in 3 segments with maximum amplitude of 0.5cm. The maximum transverse velocity amplitude is

The correct option is C 1.57m/sFrequency of the wave traveling through the stretched wire with tension T and mass per unit length m is given by(string vibrates in 3 segments):n=3/2L(√T/m)=50Hz.So, the maximum transverse velocity amplitude is:Vmax=amaxω=2πamaxn=1.57m/s.

A string of mass $0.2 k g / m$ and length $l = 0.6 m$ is fixed at both ends and stretched such that it has a tension of $80 N$ . The string vibrates in $3$ segments with maximum amplitude of $0.5 c m$ . The maximum transverse velocity amplitude is