A) \[50,\text{ }40\pi \]
B) \[100,\text{ 4}0\pi \]
C) \[100,\text{ 2}0\pi \]
D) \[100,\text{ 8}0\pi \]
Correct Answer: B
Solution :
: \[y=10sin(4\pi t-0.02\pi x)\] Standard equation is given by \[y=a\sin \left( \frac{2\pi vt}{\lambda }-\frac{2\pi }{\lambda }x \right)=a\sin (\omega t-kx)\] \[\therefore \] \[a=10\]cm ...(i) \[\frac{2\pi }{\lambda }=0.02\pi \] or \[\lambda =100\]cm ...(ii) \[\omega =4\pi \] \[\therefore \]Maximum velocity\[=a\omega =10\times 4\pi \] \[=40\pi \]cm/sec ...(iii) Hence, from (ii) and (iii), we have \[\lambda =100\text{ }cm\] \[{{v}_{m}}=100\pi \] cm/sec.You need to login to perform this action.
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