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The displacement of a particle executing simple harmonic motion is given by `x=3sin(2pit+(pi)/(4))` where x is in metres and t is in seconds. The amplitude and maximum speed of the particle isA. 3m, `2pims^(-1)`B. 3m, `4pims^(-1)`C. 3m, `6pims^(-1)`D. 3m, `8pims^(-1)` |
Answer» Correct Answer - C The given equation of SHM is `x=3sin(2pit+(pi)/(4))` Compare the given equation with standard equation of SHM. `x=Asin(omegat+phi)` we get, `A=3m,omega=2pis^(-1)`. `therefore`Maximum speed, `v_(max)=Aomega=3mxx2pis^(-1)=6pims^(-1)` |
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