JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2025
- Easy
- MCQ
A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the x -direction in 0.3 sec. The crest \(P\) is at \(x=0\) at \(t=0 \mathrm{sec}\) and maximum displacement of the wave is 2 cm. Which equation correctly represents this wave ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(y=2 \cos (0.83 x-3.35 t) \mathrm{cm}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{v}=\frac{\text { distance }}{\text { time }} \\
& \mathrm{v}=\frac{12}{0.3}=4 \mathrm{~cm} / \mathrm{s} \\
& \mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi}{7.5}=\frac{4 \pi}{15}=0.83 \\
& \mathrm{v}=\frac{\omega}{\mathrm{k}} \Rightarrow \omega=\mathrm{vk}=4 \times \frac{4 \pi}{15}=3.35
\end{aligned}\)
So \(y=A \cos (k x-\omega t)\)
& \mathrm{v}=\frac{\text { distance }}{\text { time }} \\
& \mathrm{v}=\frac{12}{0.3}=4 \mathrm{~cm} / \mathrm{s} \\
& \mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi}{7.5}=\frac{4 \pi}{15}=0.83 \\
& \mathrm{v}=\frac{\omega}{\mathrm{k}} \Rightarrow \omega=\mathrm{vk}=4 \times \frac{4 \pi}{15}=3.35
\end{aligned}\)
So \(y=A \cos (k x-\omega t)\)
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