A) 25 N
B) 5 N
C) 2.5 N
D) 0.5 N
Correct Answer: D
Solution :
Force in a simple harmonic motion \[F=mf=m(-{{\omega }^{2}}x)=-m{{\omega }^{2}}A\,\sin \,\omega t\] For maximum value of F, \[\sin \,\,\omega t=1\] \[\therefore \] \[{{F}_{\max }}=m{{\omega }^{2}}A\] \[=m{{\left( \frac{2\pi }{T} \right)}^{2}}A=\frac{m\,\,4{{\pi }^{2}}A}{{{T}^{2}}}\] \[=\frac{10\times {{10}^{-3}}\times 4{{\pi }^{2}}\times 0.5}{{{\pi }^{2}}/25}=0.5N\]You need to login to perform this action.
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