JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 3)
Displacement of a wave is expressed as $x(t)=5 \cos \left(628 t+\frac{\pi}{2}\right) \mathrm{m}$. The wavelength of the wave when its velocity is $300 \mathrm{~m} / \mathrm{s}$ is :
$$(\pi=3.14)$$
0.33 m
0.5 m
3 m
5 m
Explanation
$$\begin{aligned}
& \mathrm{x}(\mathrm{t})=5 \cos \left[628 \mathrm{t}+\frac{\pi}{2}\right] \mathrm{m} \\
& \text { velocity }\left(\mathrm{v}_\omega\right)=300 \mathrm{~m} / \mathrm{s} \\
& \mathrm{v}_{\mathrm{w}}=\frac{\omega}{\mathrm{K}} \\
& 300=\frac{628}{\mathrm{~K}} \Rightarrow \mathrm{~K}=\frac{628}{300} \\
& \frac{2 \pi}{\lambda}=\frac{628}{300} \Rightarrow \lambda=\frac{2 \times 3.14 \times 300}{628} \\
& \lambda=2 \mathrm{~m}
\end{aligned}$$
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