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The period of a particle in SHM is `8s`. At `t=0` it is at the mean position. The ratio of the distances travled by it in the first and the second second isA. `3.2:1`B. `4.2:1`C. `2.4:1`D. `1.6:` |
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Answer» Correct Answer - C `t_(1)=1s, t_(2)=2s` Now,`y_(1)=asinomegaxx1=asinomega` `y_(2)=asin omegaxx2-asinomega=asin2omega-asinomega` `:. (y_(2))/(y_(1))=(sin2omega-sinomega)/(sinomega)=(sin2omega)/(sinomega)-1` `=(2sinomegacosomega)/(sinomega)-1=2cosomega-1` `=2cos((2pi)/(T))-1=2cos((2pi)/(8))-1` or `(y_(2))/(y_(1))=2cos((pi)/(4))-1=2xx(1)/(sqrt(2))-1=sqrt(2)-1` `=1.414+1=2.414~~2.4:1` |
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