Two masses m1=1kg and m2 = 0.5kg are suspended together by a massless spring of spring constant 12.5N/m.When masses are in equilibrium m1 is removed without disturbing the system . New amplitude of oscillation will be
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Let initially, only body of mass m₂ are attached to string and extension of string by this body is l ,
Then in equilibrium,
m₂g = Kl -------(1)
where k is spring constant
Now, after attaching m₁ tnen, increase in extension of string with ∆l
in equilibrium condition,
(m₁ + m₂)g = K(l + ∆l) --------(2)
solve equations (1) and (2) ,
m₁g = K∆l
∆l = m₁g/K
now, put the value of m₁ , g and K
∆l = 1 × 10/12.5 = 0.8 m = 80 cm
Hence, amplitude of oscillation is 80 cm
Then in equilibrium,
m₂g = Kl -------(1)
where k is spring constant
Now, after attaching m₁ tnen, increase in extension of string with ∆l
in equilibrium condition,
(m₁ + m₂)g = K(l + ∆l) --------(2)
solve equations (1) and (2) ,
m₁g = K∆l
∆l = m₁g/K
now, put the value of m₁ , g and K
∆l = 1 × 10/12.5 = 0.8 m = 80 cm
Hence, amplitude of oscillation is 80 cm
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