A) 0.36
B) \[0.28\]
C) \[0.61\]
D) \[0.53\]
Correct Answer: C
Solution :
The angular velocity \[\omega =3.5\text{ }red/sec\] maximum acceleration \[{{a}_{\max }}=7.5m/{{s}^{2}}\] We know for a SHM, the displacement \[x=A\sin \omega t\] \[\therefore \] \[v=\frac{dx}{dt}=A\omega \cos \omega t\] \[\therefore \] \[a=\frac{dv}{dt}=-A{{\omega }^{2}}\sin \omega t\] \[\therefore \] Maximum acceleration \[|{{a}_{\max }}|=A{{\omega }^{2}}\] Now \[A{{\omega }^{2}}=7.5\] \[\Rightarrow \] \[A=\frac{7.5}{{{\omega }^{2}}}\] \[=\frac{7.5}{{{(3.5)}^{2}}}=0.6\] \[\therefore \]Amplitude = 0.6
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