MHT CET · Physics · Waves and Sound
Two progressive waves \(\mathrm{Y}_{1}=\sin 2 \pi\left(\frac{\mathrm{t}}{0 \cdot 4}-\frac{\mathrm{x}}{4}\right)\) and \(\mathrm{Y}_{2}=\sin 2 \pi\left(\frac{\mathrm{t}}{0 \cdot 4}+\frac{\mathrm{x}}{4}\right)\) superpose to form a standing wave. \(\mathrm{x}, \mathrm{Y}_{1}\) and \(\mathrm{Y}_{2}\) are in SI system. Amplitude of the particle at \(\mathrm{x}=0.5 \mathrm{~m}\) is \(\left[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]\)
- A \(2 \sqrt{2} \mathrm{~m}\)
- B \(2 \mathrm{~m}\)
- C \(\sqrt{2} \mathrm{~m}\)
- D \(\frac{1}{\sqrt{2}} \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{2} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
The stationary wave is given by
\(y=2 \cos \frac{2 \pi x}{\lambda} \sin \frac{2 \pi t}{T}\)
where \(\lambda=4 \mathrm{~m}\) and \(\mathrm{T}=0.4 \mathrm{~s}\)
Amplitude \(A=2 \cos \frac{2 \pi x}{\lambda}\)
At \(x=0.5 \mathrm{~m}, A=2 \cos \frac{2 \pi \times 0.5}{4}\)
\(A=2 \cos \frac{\pi}{4}=2 \times \frac{1}{\sqrt{2}}=\sqrt{2} m\)
\(y=2 \cos \frac{2 \pi x}{\lambda} \sin \frac{2 \pi t}{T}\)
where \(\lambda=4 \mathrm{~m}\) and \(\mathrm{T}=0.4 \mathrm{~s}\)
Amplitude \(A=2 \cos \frac{2 \pi x}{\lambda}\)
At \(x=0.5 \mathrm{~m}, A=2 \cos \frac{2 \pi \times 0.5}{4}\)
\(A=2 \cos \frac{\pi}{4}=2 \times \frac{1}{\sqrt{2}}=\sqrt{2} m\)
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