InterviewSolution
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InterviewSolution
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A particle executes simple harmotic motion about the point `x=0`. At time `t=0`, it has displacement `x=2cm` and zero velocity. If the frequency of motion is `0.25 s^(-1)`, find (a) the period, (b) angular frequency, ( c) the amplitude, (d) maximum speed, (e) the displacement from the mean position at `t=1 s`and (f) the velocity at `t=1 s`. |
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Answer» (i) Period , `T=(1)/(f)=(1)/(0.25 s^(-1))=4 s` (ii) Angular frequency, `omega=(2pi)/(T)=(2pi)/(4)=(pi)/(2)"rad"s^(-1)=1.57 "rad"s^(-1)` (iii) Amplitude is the maximum displacement from mean position. Hence, A=2-0=2 cm. (iv) Maximum speed, `v_("max")=Aomega=2.(pi)/(2)=pi cms^(-1)=3.14 cms^(-1)` (v) The displacement is given by `x=A"sin"(omegat+phi)` Initially at t=0, x=2 cm ,then `2=2"sin"phi` or `"sin"phi=1="sin"90^(@)` or `phi=90^(@)` Now at t=3s `x=2"sin"((pi)/(2)xx3+(pi)/(2))=0` (vi) Velocity at x=0 is `v_("max")`. `therefore v_("max")=omegaA=((pi)/(2))(2)=3.14 cms^(-1)` |
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