A) Travelling with a velocity of 30 m/sec in the negative X direction
B) Of wavelength p metre
C) Of frequency 30/p Hz
D) Of amplitude \[{{10}^{4}}\]metre travelling along the negative X direction
Correct Answer: A
Solution :
On comparing the given equation with \[y=a\sin (\omega \,t+kx),\] it is clear that wave is travelling in negative x-direction. It's amplitude a = 104 m and w = 60, k = 2. Hence frequency \[n=\frac{\omega }{2\pi }=\frac{60}{2\pi }=\frac{30}{\pi }Hz\] \[k=\frac{2\pi }{\lambda }=2\] Þ \[\lambda =\pi \,m\] and \[v=\frac{\omega }{k}=\frac{60}{2}=30\,m/s\]
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