wave_mechanics mock_test_waves_and_acoustics Question 17
Question: The equation of a wave on a string of linear mass density 0.04 kg/m is given by$ y=0.02(m)sin[ 2\pi ( \frac{t}{0.04(s)}-\frac{x}{0.50(m)} ) ] $ The tension in the string is
Options:
A) 4.0 N
B) 12.5 N
C) 0.5 N
D) 6.25 N
Show Answer
Answer:
Correct Answer: D
Solution:
[d] The given equation of a wave is $ y=0.02\sin [ 2\pi ( \frac{t}{0.04}-\frac{x}{0.50} ) ] $ Compare it with the standard wave equation $ y=A\sin (\omega t-kx) $ We get $ \omega =\frac{2\pi }{0.04}\text{rad}{{s}^{-1}};k=\frac{2\pi }{0.5}\text{rad}{{m}^{-1}} $ Wave velocity,$ v=\frac{\omega }{k}=\frac{( 2\pi /0.04 )}{( 2\pi /0.5 )}m{{s}^{-1}} $ …(i) Also$ v=\sqrt{\frac{T}{\mu }} $ …(ii) Where T is the tension in the string and $ \mu $ .
is the linear mass density here, linear mass density.$ \mu =0.04kg{{m}^{-1}} $ Equating equations (i) and (ii), we get $ \frac{\omega }{k}=\sqrt{\frac{T}{\mu }} $ or $ T=\frac{\mu {{\omega }^{2}}}{k^{2}} $
$ T=\frac{0.04\times {{( \frac{2\pi }{0.04} )}^{2}}}{{{( \frac{2\pi }{0.05} )}^{2}}}=6.25N $
BETA
