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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
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B
2
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C
12
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D
2+1
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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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