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  4. The maximum velocity of a particle executing SHM is 'v'. If the amplitude is doubled and the time period of oscillation decreased to 1/3 of its original value, the maximum velocity becomes :-

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Q. The maximum velocity of a particle executing SHM is '$v$'. If the amplitude is doubled and the time period of oscillation decreased to $1/3$ of its original value, the maximum velocity becomes :-

Solution:

$V _{\max _{1}}=\omega A =\left(\frac{2 \pi}{ T }\right) A = V$
$V _{\max _{2}}=\left(\frac{2 \pi}{ T / 3}\right)(2 A )=6\left(\frac{2 \pi}{ T }\right) A =6\, V$