A wave is given by $y = 3\sin 2\pi \left( {\frac{t}{{0.04}} - \frac{x}{{0.01}}} \right)$, where $y$ is in $cm$. Frequency of wave and maximum acceleration of particle will be
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(d) Comparing the given equation with standard equation
$y = a\sin 2\pi \,\left( {\frac{t}{T} - \frac{x}{\lambda }} \right)$==> $T = 0.04 sec$
==> $v = \frac{1}{T} = 25Hz$
Also ${(A)_{\max }} = {\omega ^2}a = {\left( {\frac{{2\pi }}{T}} \right)^2} \times a = {\left( {\frac{{2\pi }}{{0.04}}} \right)^2} \times 3$
$=7.4 \times 10^4 cm/sec^2.$
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