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
B
1:2
C
1:3
D
2:1

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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