JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2025
- Easy
- MCQ
The equation of a transverse wave travelling along a string is \(y(x, t)=4.0 \sin \left[20 \times 10^{-3} x+600 t\right] \mathrm{mm}\), where \(x\) is in mm and \(t\) is in second. The velocity of the wave is :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(-30 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& k=20 \times 10^{-3} \mathrm{~mm}^{-1}=20 \mathrm{~m}^{-1} \\
& w=600 \mathrm{~s}^{-1} \\
& v=\frac{W}{k}=\frac{600}{20}=30 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
and \(x \& t\) carry same sign
Therefore \(\mathrm{v}=-30 \mathrm{~m} / \mathrm{s}\)
& k=20 \times 10^{-3} \mathrm{~mm}^{-1}=20 \mathrm{~m}^{-1} \\
& w=600 \mathrm{~s}^{-1} \\
& v=\frac{W}{k}=\frac{600}{20}=30 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
and \(x \& t\) carry same sign
Therefore \(\mathrm{v}=-30 \mathrm{~m} / \mathrm{s}\)
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