Question

two masses, each having a value of m, are vibrating vertically on a spring with a hookes law constant, k. at the lowest point of the vibration, one of the masses falls off, so that now the total mass is m instead of 2m. comparing the new vibrational motion to the original vibrational motion: 1) how is the period of vibration different, if at all? explain you reasoning. 2) how is the maximum acceleration different, if at all? explain your reasoning. 3) how is the maximum velocity different, if at all? explain your reasoning. quantitative comparisons are preferred: i.e., one - third as large or twice as fast.

Explanation:

Step1: Period of vibration

The formula for the period of a spring - mass system is \(T = 2\pi\sqrt{\frac{m}{k}}\).
Originally, \(m = 2M\), so \(T_{1}=2\pi\sqrt{\frac{2M}{k}}\).
After one mass falls off, \(m = M\), so \(T_{2}=2\pi\sqrt{\frac{M}{k}}\).
Dividing \(T_{2}\) by \(T_{1}\): \(\frac{T_{2}}{T_{1}}=\frac{2\pi\sqrt{\frac{M}{k}}}{2\pi\sqrt{\frac{2M}{k}}}=\frac{1}{\sqrt{2}}\).

Step2: Maximum acceleration

The maximum force on the system originally is \(F_{1}=kA_{1}\) (from Hooke's law \(F = kx\), at maximum displacement \(x = A\)), and \(F_{1}=2Ma_{1,\text{max}}\) (from \(F = ma\)). So \(a_{1,\text{max}}=\frac{kA_{1}}{2M}\).
At the lowest point, the force \(F = kA\) (the displacement from the new equilibrium position). After the mass falls off, the force \(F_{2}=kA_{2}\), and \(F_{2}=Ma_{2,\text{max}}\).
The equilibrium position changes. Let the original equilibrium position be \(x_{01}=\frac{2Mg}{k}\), and the new equilibrium position be \(x_{02}=\frac{Mg}{k}\). At the lowest - point (before mass falls), the displacement from the original equilibrium \(A_{1}\) satisfies \(k(A_{1}+x_{01})-2Mg = 2Ma_{1,\text{max}}\) (using \(F = ma\), \(F = k(A_{1}+x_{01})-2Mg\)). After the mass falls, the displacement from the new equilibrium \(A_{2}\): \(k(A_{2}+x_{02})-Mg = Ma_{2,\text{max}}\). Since at the moment the mass falls, the spring is at the same physical position. Let \(x\) be the displacement from the natural length of the spring. \(x=x_{01}+A_{1}=x_{02}+A_{2}\). Substituting \(x_{01}=\frac{2Mg}{k}\) and \(x_{02}=\frac{Mg}{k}\), we get \(A_{2}=A_{1}+\frac{Mg}{k}\).
Another way: Using energy. The potential energy at the lowest point \(U=\frac{1}{2}kA^{2}\). The force \(F = kA\). Originally \(F_{1}=kA_{1}\) and \(F_{1}=2Ma_{1,\text{max}}\), after \(F_{2}=kA_{2}\) and \(F_{2}=Ma_{2,\text{max}}\). From the equilibrium change, \(A_{2} = 2A_{1}\). So \(a_{2,\text{max}}=\frac{kA_{2}}{M}=2\frac{kA_{1}}{M}\), and \(a_{1,\text{max}}=\frac{kA_{1}}{2M}\). So \(a_{2,\text{max}} = 4a_{1,\text{max}}\).

Step3: Maximum velocity

The maximum velocity of a simple harmonic oscillator is \(v_{\text{max}}=\omega A\), where \(\omega=\sqrt{\frac{k}{m}}\).
Originally, \(\omega_{1}=\sqrt{\frac{k}{2M}}\) and \(v_{1,\text{max}}=\omega_{1}A_{1}=\sqrt{\frac{k}{2M}}A_{1}\).
After, \(\omega_{2}=\sqrt{\frac{k}{M}}\) and \(A_{2} = 2A_{1}\), so \(v_{2,\text{max}}=\omega_{2}A_{2}=\sqrt{\frac{k}{M}}\times2A_{1}\).
\(\frac{v_{2,\text{max}}}{v_{1,\text{max}}}=\frac{\sqrt{\frac{k}{M}}\times2A_{1}}{\sqrt{\frac{k}{2M}}A_{1}} = 2\sqrt{2}\).

Answer:

1. The new period \(T_{2}\) is \(\frac{1}{\sqrt{2}}\) times the original period \(T_{1}\).
2. The new maximum acceleration \(a_{2,\text{max}}\) is \(4\) times the original maximum acceleration \(a_{1,\text{max}}\).
3. The new maximum velocity \(v_{2,\text{max}}\) is \(2\sqrt{2}\) times the original maximum velocity \(v_{1,\text{max}}\).