Question

# Time period of a particle executing SHM is 8 sec. At t=0 it is at the mean position. The ratio of the distance covered by the particle in the 1st second to the 2nd second is:

A
12+1
B
2
C
12
D
2+1

Solution

## The correct option is D $${\displaystyle\sqrt{2}+1}$$$$x=A\sin { \omega t } \\ x\left( t=1s \right) =A\sin { \left( \dfrac { 2\pi }{ T } t \right) } =A\sin { \left( \dfrac { 2\pi }{ 8 } \right) } =A\sin { \left( \dfrac { \pi }{ 4 } \right) } =\dfrac { A }{ \sqrt { 2 } } \\ x\left( t=2s \right) =A\sin { \left( \dfrac { 2\pi }{ T } t \right) } =A\sin { \left( \dfrac { 2\pi }{ 8 } \times 2 \right) } =A\sin { \left( \dfrac { \pi }{ 2 } \right) } =A$$And the required ratio is:$$\dfrac { x\left( t=1s \right) }{ x\left( t=2s \right) -x\left( t=1s \right) } =\dfrac { \dfrac { A }{ \sqrt { 2 } } }{ A-\dfrac { A }{ \sqrt { 2 } } } =\dfrac { 1 }{ \sqrt { 2 } -1 } =\sqrt { 2 } +1$$Physics

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