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A platform moves up and down in SHM, with amplitude 0.050 m. Resting on top of the platform is a block of wood. What is the shortest period of motion for the platform so that the block will remain in constant contact with it?

Respuesta :

Answer:

Shortest time period of S.H.M will be equal to 0.1418 sec        

Explanation:

We have given amplitude of S.H.M is A = 0.050 m

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex] ( This will be maximum acceleration )We know that maximum  acceleration is equal to [tex]\omega ^2A[/tex]

So [tex]g=\omega ^2A[/tex]

[tex]9.8=\omega ^2\times 0.050[/tex]

[tex]\omega =44.271rad/sec[/tex]

We have to find the shortest time period

Now angular velocity is equal to [tex]\omega =\frac{2\pi }{T}[/tex]

So time period [tex]T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{44.271}=0.1418sec[/tex]

So shortest time period of S.H.M will be equal to 0.1418 sec

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