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MakcuM [25]
2 years ago
11

A 5.0 m length of rope, with a mass of 0.52 kg, is pulled taut with a tension of 46 N. Find the speed of waves on the rope

Physics
1 answer:
aleksandrvk [35]2 years ago
4 0

Answer:

Speed of waves on the rope is 21 m/s

Explanation:

Length of the rope (l) = 5.0 m

Mass of the rope (m) = 0.52 kg

Tension in the rope (T) = 46 N

Formula of speed of waves on the rope:

\bold{v = \sqrt{\dfrac{T}{\mu}}}

\mu = Mass per unit length of the rope (m/l)

By substituting the values in the formula we get:

\implies  \rm v = \sqrt{\dfrac{T}{ \dfrac{m}{l} }} \\  \\  \implies  \rm v = \sqrt{\dfrac{Tl}{m}} \\  \\ \implies  \rm v =  \sqrt{ \dfrac{46 \times 5}{0.52} }  \\  \\ \implies  \rm v =  \sqrt{ \dfrac{230}{0.52} }  \\  \\ \implies  \rm v =  \sqrt{442.3}  \\  \\ \implies  \rm v = 21 \: m {s}^{ - 1}

Speed of waves on the rope (v) = 21 m/s

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d=\frac{m\lambda }{\sin \theta _m}-----1

Position of bright fringe is given by

y=\tan \theta _m\cdot L

\tan \theta _m=\frac{y}{L}

\theta _m=\tan^{-1}(\frac{y}{L})

Put the value of \theta _m  in eq. 1

d=\frac{m\lambda }{\sin (\tan^{-1}(\frac{y}{L}))}

Therefore N=d^{-1}

N=\frac{\sin (\tan^{-1}(\frac{y}{L}))}{m\lambda }

for m=1

N=\frac{\sin (\tan^{-1}(\frac{1.9\times 10^{-2}}{0.8}))}{1\times 590\times 10^{-9}}

N=40243\ line/m

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Answer:

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