MHT CET · Physics · Waves and Sound
The equation of simple harmonic wave produced in the string under tension \(0.4 \mathrm{~N}\) is given by \(\mathrm{y}=4 \sin (3 \mathrm{x}+60 \mathrm{t}) \mathrm{m}\). The mass per unit length of the string is
- A \(10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}\)
- B \(10^{-5} \mathrm{~kg} \mathrm{~m}^{-1}\)
- C \(10^{-3} \mathrm{~g} \mathrm{~cm}^{-1}\)
- D \(10^{-5} \mathrm{~g} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(A) \(10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}\)
Step-by-step Solution
Detailed explanation
The standard equation of a wave can be written as
\(
\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})
\)
Speed of wave \(\mathrm{V}=\frac{\omega}{\mathrm{k}}=\frac{60}{3}=20 \mathrm{~m} / \mathrm{s}\)
Also, \(V=\sqrt{\frac{T}{m}}\)
\(
\therefore \mathrm{m}=\frac{\mathrm{T}}{\mathrm{V}^2}=\frac{0.4}{400}=10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}
\)
\(
\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})
\)
Speed of wave \(\mathrm{V}=\frac{\omega}{\mathrm{k}}=\frac{60}{3}=20 \mathrm{~m} / \mathrm{s}\)
Also, \(V=\sqrt{\frac{T}{m}}\)
\(
\therefore \mathrm{m}=\frac{\mathrm{T}}{\mathrm{V}^2}=\frac{0.4}{400}=10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}
\)
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