Solveeit Logo
Login

Question

Question: The transverse displacement of a string clamped at its both ends is given by y(x, t) = 0.06 sin cos ...

The transverse displacement of a string clamped at its both ends is given by y(x, t) = 0.06 sin cos (120pt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3 × 10–2 kg. The tension in the string is

A

324 N

B

648 N

C

832 N

D

972 N

Answer

648 N

Explanation

Solution

The given equations is

y(x,t)=0.06sin⁡(2x3x)cos⁡(120πt)y(x,t) = 0.06\sin\left( \frac{2x}{3}x \right)\cos(120\pi t)

Compare it with y(x,t) = 2a sin kx cos⁡ωt\cos\omega t

We get

k=2π3,or2πλ=2π3orλ=3mk = \frac{2\pi}{3},or\frac{2\pi}{\lambda} = \frac{2\pi}{3}or\lambda = 3m

And ω=120πor2πυ=120π\omega = 120\pi or2\pi\upsilon = 120\pi

Or υ=60Hz=60s−1\upsilon = 60Hz = 60s^{- 1}

Velocity of wave,

v=υλ=(60s−1)(3m)=180ms−1v = \upsilon\lambda = (60s^{- 1})(3m) = 180ms^{- 1}

Mass per unit length of the string.

μ=3×10−2kg1.5m=2×10−2kgm−1\mu = \frac{3 \times 10^{- 2}kg}{1.5m} = 2 \times 10^{- 2}kgm^{- 1}

Velocity of transverse wave in the string.

v=Tμorv2=TμorT=v2μv = \sqrt{\frac{T}{\mu}}orv^{2} = \frac{T}{\mu}orT = v^{2}\mu

T=(180ms−1)2(2×10−2kgm−1)=648NT = (180ms^{- 1})^{2}(2 \times 10^{- 2}kgm^{- 1}) = 648N