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CAF123
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Homework Statement
A box of mass m is released from rest at ##x_o = 0## with the two springs ##k## and ##2k## unstretched. Find the distance that the mass falls before rebounding. What maximum velocity does it obtain?
Homework Equations
Hookes law, gravitational force.
The Attempt at a Solution
Suppose the distance the top spring is stretched is ##x_1## and that of the lower spring, ##x_2##. Then define ##z = x_1 + x_2##. At equilibrium, ##mg - 2kx_2 = 0## and ##2kx_2 = kx_1##. (There is an upwards force of ##kx_1## from the displacement of the upper spring and a downwards force of ##kx_2## since the extension of the lower spring causes a pull on the upper spring and by virtue of NIII, there is a corresponding reaction force). This means ##x_1 = 2x_2## so that ##z = 3x_2## and finally ##mg - 2/3 k z = 0## at equilibrium. So the system is equivalent to one with a spring constant of 2/3 and a displacement of the mass a distance z at equilibrium. In general, we can then write that the mass satisfies ##mg - (2/3) k Z = m dv/dt##, where Z is some extension of the equivalent system with constant 2/3. Write this as $$g - \frac{2}{3} \frac{k}{m} Z = v\frac{dv}{dZ} \Rightarrow Zg - \frac{2}{3}\frac{k}{m} \frac{Z^2}{2} = \frac{v^2}{2}$$ so that $$v = \sqrt{2gZ - \frac{2}{3}\frac{k}{m}Z^2}$$ The distance the mass falls before rebounding is given by the first instant (other than at the point of release) when the velocity is zero. This implies ##Z_{\text{before rebound}} = 3mg/k## while maximum velocity is obtained from ##dv/dZ = 0 \rightarrow Z = 3/2 (mg/k ).##. Is this correct?