InterviewSolution
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InterviewSolution
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A particle executing simple harmonic motion has angular frequency `6.28 s^(-1)` and amplitude `10 cm`. Find `(a)` the time period, `(b)` the maximum speed, `(c)` the maximum acceleration, `(d)` the speed when the displacement is `6 cm` from the mean position, `(e)` the speed at `t = 1//6 s` assuming that the motion starts from rest at `t = 0`. |
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Answer» a. Time period `=(2pi)/6.28s=1s` b. Maimum speed `=Aomega=(0.1m)(6.28m^-1)` `=0.628ms^-1` c. Maximum acceleration `=Aomega^2` ltbr.gt `=(0.1m)(6.28s^-1)^2` =4ms^-2` d. `v=omegasqrt(A^2-x^2)=(6.28s^-1)sqrt((10cm)^2-(6cm)^2)` `=50.2cms^-1` c. At t=0 the velocity is zero i.e. the particle is at an extreme. The equation for displacement may be written as `x=Acosomegat` The velocity is ` v=-Aomegasinomegat`. At `t=1/6s, v=-(0.1m)(6.28s^-1)sin(6.28/6)` `=(-0.628ms^-1)sinpi/3` `=-54.4cms^-1` |
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