MHT CET · Physics · Waves and Sound
A transverse wave given by \(\mathrm{y}=2 \sin (0.01 \mathrm{x}+30 \mathrm{t})\) moves on a stretched string from one end to another end in 0.5 second. If ' \(x\) ' and ' \(y\) ' are in \(\mathrm{cm}\) and ' \(\mathrm{t}\) ' is in second, then the length of the string is
- A 6m
- B 9m
- C 12m
- D 15m
Answer & Solution
Correct Answer
(D) 15m
Step-by-step Solution
Detailed explanation
Comparing with the standard equation
\(
\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})
\)
we get \(\mathrm{k}=0.01 \mathrm{rad} / \mathrm{cm}\) and \(\omega=30 \mathrm{rad} / \mathrm{s}\)
speed of the wave, \(v=\frac{\omega}{\mathrm{k}}=\frac{30}{0.01}=300 \mathrm{~cm} / \mathrm{s}=30 \mathrm{~m} / \mathrm{s}\)
\(\therefore\) Distance \(=\mathrm{vt}=30 \times 0.5=15 \mathrm{~m}\)
\(
\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})
\)
we get \(\mathrm{k}=0.01 \mathrm{rad} / \mathrm{cm}\) and \(\omega=30 \mathrm{rad} / \mathrm{s}\)
speed of the wave, \(v=\frac{\omega}{\mathrm{k}}=\frac{30}{0.01}=300 \mathrm{~cm} / \mathrm{s}=30 \mathrm{~m} / \mathrm{s}\)
\(\therefore\) Distance \(=\mathrm{vt}=30 \times 0.5=15 \mathrm{~m}\)
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