JEE PHYSICS MCQ: WAVES

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Application +4 | -1

The transverse displacement of a stretched string which is connected between two rigid supports is given by

y = 0.06\sin \left( {\frac{{2\pi x}}{5} - 80\pi t} \right)

where x and y are in metre and t in sec. Length of string is 1.5 m and its mass is

3 \times {10^{ - 2}}\,kg.

Then tension in string will be:-

View answer and explanation

Correct option: C

800 N

Option C (800 N) is correct because the tension in the string is calculated using the formula

T = \mu v^2 = \mu (\frac{\omega}{k})^2

, where

\mu

is the linear mass density,

\omega

is the angular frequency, and k is the wave number.

More Questions from WAVES

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3\,\,kHz

0.5

f\mathrm{ }Hz

V\mathrm{ }m/sec

0.2\,\,V

y = \frac{3}{{{{\left( {2x + 3t} \right)}^2} + 4}},

(– \mathrm{ve}) \,x

y = f(kx +\omega t),

X

0.01\,\,s

5 \mathrm{~m}

245 \mathrm{~g}

1 \mathrm{~kg}

g=980 \mathrm{~cm} \mathrm{~s}^{-2}

T

V