TS EAMCET · Physics · Waves and Sound
The equation of a transverse wave propagating on a stretched string is given by \(y=3 \sin (4 x+200 \mathrm{t})\), where \(x\) and \(y\) are in metre and the time ' t ' is in second. If the tension applied to the string is 500 N, the linear density of the string is
- A \(0.25 \mathrm{~kg} \mathrm{~m}^{-1}\)
- B \(0.4 \mathrm{~kg} \mathrm{~m}^{-1}\)
- C \(0.2 \mathrm{~kg} \mathrm{~m}^{-1}\)
- D \(0.1 \mathrm{~kg} \mathrm{~m}^{-1}\)
Answer & Solution
Correct Answer
(C) \(0.2 \mathrm{~kg} \mathrm{~m}^{-1}\)
Step-by-step Solution
Detailed explanation
\(k = 4 \ \mathrm{m}^{-1}\), \(\omega = 200 \ \mathrm{rad/s}\) \(v = \frac{\omega}{k} = \frac{200}{4} = 50 \ \mathrm{m/s}\) \(\mu = \frac{T}{v^2} = \frac{500}{(50)^2} = \frac{500}{2500} = 0.2 \ \mathrm{kg \ m^{-1}}\)
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