Waves

JEE Main level, single correct, medium

26 of 142

Question 26 of 142

Displacement of a wave is expressed as x(t)=5cos⁡(628t+π2)mx(t)=5 \cos \left(628 t+\frac{\pi}{2}\right) m. The wavelength of the wave when its velocity is 300 m/s300 \mathrm{~m} / \mathrm{s} is :

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Answer: (B) 3 m

x(t)=5cos⁡[628t+π2]m velocity (vω)=300 m/svw=ωK300=628 K⇒ K=6283002πλ=628300⇒λ=2×3.14×300628λ=2 m\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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