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A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple harmonic motion with a period of 1s. If that same mass–spring system is displaced from equilibrium by 10cm instead, what will its period be in this case?(A) 0.5s(B) 2s(C) 1s(D) 1.4s

Respuesta :

Answer:

T= 1 s

Explanation:

Given that

When x=  cm ,T= 1

we know that time period of spring mas system given as

[tex]T=2\pi \sqrt{\dfrac{m}{k}}[/tex]

T= Time period

m= mass

k=spring constant

So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

T= 1 s

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