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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
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B
52π
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C
4π5
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D
2π3
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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 period
Now, 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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