Question

# A particle executes linear simple harmonic motion with an amplitude of $$3\ cm$$. When the particle is at $$2\ cm$$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

A
5π
B
52π
C
4π5
D
2π3

Solution

## The correct option is C $$\dfrac {4\pi}{\sqrt {5}}$$Velocity of particle executing S.H.M $$v=\omega \sqrt { { a }^{ 2 }-{ x }^{ 2 } }$$ and acceleration, $$a=-{ \omega }^{ 2 }x$$ $$x$$= displacement at any time interval$$a$$ = amplitude$$=3cm$$$$\omega$$= angular frequency, $$=\cfrac { 2\pi }{ T }$$$$T$$=Time periodNow, at $$x=2cm$$ $$v=|a|$$ $$\Rightarrow { \omega }^{ 2 }\times 2=\omega \sqrt { { 3 }^{ 2 }-{ 2 }^{ 2 } }$$$$\Rightarrow { \omega }=\cfrac { \sqrt { 5 } }{ 2 }$$$$\Rightarrow \cfrac { 2\pi }{ T } =\cfrac { \sqrt { 5 } }{ 2 }$$$$\Rightarrow T=\cfrac { 4\pi }{ \sqrt { 5 } }$$Physics

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