Question

Maximum velocity of a particle performing linear shm is 0.16 metre per second if its maximum acceleration is 0.64 metre per second square calculate its period​

Answer 1

$\mathfrak{ \huge{ \red{ \underline{ \underline{ANSWER}}}}} \\ \\ \star \textrm{ \red{Given}} \\ \\ \implies \sf \: maximum \: vecity \: of \: particle = 0.16 \: \frac{m}{s} \\ \\ \implies \sf \: maximum \: acceleration \: of \: particle = 0.64 \: \frac{m}{ {s}^{2} } \\ \\ \star \textrm{ \red{To \: Find}} \\ \\ \implies \sf \: time \: period \: of \: particle \\ \\ \star \: \textrm{ \red{Formula}} \\ \\ \implies \sf \: maximum \: velocity \: of \: particle \: is \: given \: as \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{\pink{ \bold{V{ \tiny{max}} = A \omega}}}} \\ \\ \implies \sf \: maximum \: acceleration \: of \: particle \: is \: given \: by \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bold{ \sf{ \pink{a{ \tiny{max}} = - { \omega}^{2} A}}}} \\ \\ \implies \sf \: negative \: sign \: represents \: that \: direction \: of \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: acceleration \: is \: opposite \: to \: motion \\ \\ \implies \sf \: Time \: period \: (T)\: of \: particle \: is \: given \: as \\ \\ \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bold{ \sf{ \pink{ \frac{a{ \tiny{max}}}{V{ \tiny{max}}} = \frac{ { \omega}^{2}A }{ A\omega} = \omega = \frac{2 \pi}{T} }}}} \\ \\ \star \: \textrm{ \red{Calculation}} \\ \\ \implies \sf \: \frac{a{ \tiny{max}}}{V{ \tiny{max}}} = \frac{2 \pi}{T} \\ \\ \implies \sf \: T = \frac{2 \pi \times V{ \tiny{max}}}{a{ \tiny{max}}} = \frac{2 \times 3.14 \times 0.16}{0.64} = 1.57 \: s \\ \\ \\ \huge{ \blue{\star}} \: \: \: \boxed{ \boxed{ \bold{ \sf{ \green{Time \: period = 1.57 \: s}}}}}$