Two masses (m_1) and (m_2) are suspended together by a massless spring of spring constant (k). When the masses are in equilibrium, (m_1) is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of (m_2).
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Correct Answer – A
Let `l_(1)` and `l_(2)` be the extension produced by masses `m_(1)` and `m_(2)` respectively.
Then, `(m_(1) +m_(2))g=k(l_(1)+l_(2))…(i)`
After `m_(1)` is removed , `m_(2)g=kl_(2)`…(ii)
Eqs. (i) -(ii) gives `m_(1)g=kl_(1)` or `l_(1)=(m_(1)g)/(k)`
Now, angular frequency of `m_(2), omega=(2pi)/(T)=2pisqrt((k)/(m_(2)))`
Correct Answer – B
Correct Answer – A::B.
When masss (m_(1)) is removed then equilibrum will get disturbed. There will be a restoring forec is the upward direction. The body will undergo S.H.M. now.
Let x_(1) be the extension of the spring when (m_(1)_m_(2)) are suspended and x_(2) be the extension of the sppring when `m_(1)` is remved.
:. `kx_(1) =(m_(1)+m_(2))g or `(x_91) =(m_(1)+_(m2))g`
and `kx_(2) =m_(2)g` or `x_(2) =(m_(2)g)/k`
Amplitude of mean position. Restoring force will be
or `A =((m_(1_+m_(2))g-m_(2)g)/k =(m_(1))/k`
Let at any instant the mass `m_(2)` be having a displacement x from the mean position. Restoring force will be
`F=-kx` or `m_(2)a=-kx` rArr `a k/(m_(2)x`
Comparing this with `a=-omega^(2)x`,
we get `omega^(2)=k/(m^(2) rArr omega=sqrtk/m_(2)`.