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Question
Find the time period of mass M when displaced from its equilibrium position and then released for the system shown in figure.
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Solution
For the calculation purpose. in this situation, we will neglect gravity because it is constant throughout and will not affect the net restoring force.
Let in the equilibrium position, the spring has extended by an amount x0.
Now, if the mass is given a further displacement downwards by an amount of x. The string and spring both should increase in length by x.
But. string is inextensible, hence the spring alone will contribute the total extension x + x = 2x, to lower the mass down by x from the initial equilibrium mean position x0. So, net extension in the spring (= 2x + x0)
Now force on the mass before bullying (in the x0 extension case)
F = 2T
But T = kx0 ......[Where k is spring constant]
∴ F = 2kx0 ......(i)
When the mass is lowered further by x,
F' = 2T'
But new spring length = (2x + x0)
∴ F' = 2k(2x + x0) ......(ii)
Restoring force on the system
`F_"restoring" = - [F^' - F]`
Using equations (i) and (ii), we get
`F_"restoring" = -[2k(2x + x_0) - 2kx_0]`
= `- [2 xx 2kx + 2kx_0 - 2kx_0]`
= `- 4kx`
or Ma = `- 4kx`
Where, a = acceleration .....(As, F = ma)
⇒ a = `- ((4k)/M)x`
k, M is constant.
∴ a ∝ – x
Hence, the motion is S.H.M
Comparing the above acceleration expression with standard SHM equation a = – ω2x, we get
`ω^2 = (4k)/M`
⇒ `ω = sqrt((4K)/M)`
∴ Time period T = `(2pi)/ω = (2pi)/sqrt((4K)/M) = 2pi sqrt(M/(4k))`
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