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Question

A particle executes SHM of type $$x=asin\omega t$$. It takes time $$t_1$$ from $$x=0$$ to $$x=\dfrac{a}{2}$$  and $$t_2$$ from $$x=\dfrac{a}{2}$$ $$t_2$$ from $$x=\dfrac{a}{2}$$ to $$x=a$$. The ratio of $$t_1:t_2$$ will be:


A
1:1
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B
1:2
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C
1:3
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D
2:1
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Solution

The correct option is B $$1:2$$
By symmetry the time period from $$a=0$$ to $$a=a$$ is $$T/4$$. Let eliminate the options the time taken at extreme i.e. at $$n=a$$ the time taken is more at extreme than at mid position so $${ t }_{ 2 }>{ t }_{ 1 }$$
$$\dfrac { a }{ 2 } =asinw{ t }_{ 1 }\Rightarrow { wt }_{ 1 }=\dfrac { \pi  }{ 6 } \Rightarrow $$
$${ t }_{ 1 }=\dfrac { \pi  }{ 6 } \times \dfrac { T }{ 2\pi  } =\dfrac { T }{ 12 } $$
$${ t }_{ 2 }=\dfrac { T }{ 4 } -\dfrac { T }{ 12 } =\dfrac { T }{ 6 } $$
$$\dfrac { { t }_{ 2 } }{ { t }_{ 1 } } =\dfrac { 1 }{ 2 } $$

Physics

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