# In simple harmonic motion of a spring, how does increasing the mass affect the amplitude and what equations do I use to prove it?

## I would assume the amplitude would decrease if mass was added, but I don't know how to prove it.

May 18, 2018

See below for a possible way to approach this.

#### Explanation:

Maybe take an existing system:

• $m \ddot{x} + k x = 0$

• $x = A \sin \omega t , q \quad \omega = \sqrt{\frac{k}{m}}$

• $\dot{x} = \omega A \cos \omega t$

At displacement $\xi$ and time $\tau$ the mass suddenly becomes $M$.

• $m \dot{x} = M \dot{\xi}$, where $\dot{\xi}$ is the new velocity $= \frac{m}{M} \dot{x} \left(\tau\right)$

We then treat it as a new system, extension is same so force in spring preserved. The spring is now trying to pull something with different mass back to the equilibrium but that thing has lost or gained velocity due to conservation of momentum.

$M \ddot{\xi} + k \xi = 0$

Which will solve as:

• $\xi = P \cos \Omega t + Q \sin \Omega t , q \quad \Omega = \sqrt{\frac{k}{M}}$

• $\dot{\xi} = - \Omega P \sin \Omega t + \Omega Q \cos \Omega t$

Start the clock running again, with these IV's:

• ${\dot{\xi}}_{o} = \frac{\omega A m}{M} \cos \omega \tau = \Omega Q \implies Q = \frac{\omega A m}{\Omega M} \cos \omega \tau$

• ${\xi}_{o} = A \sin \omega \tau = P$

So our new governing equation is:

• $\xi = A \sin \omega \tau \cos \Omega t + \frac{\omega A m}{\Omega M} \cos \omega \tau \sin \Omega t$

To find the new amplitude, re-write as:

• $\xi = \alpha \cos \Omega t + \beta \sin \Omega t$

• $\xi = \sqrt{{\alpha}^{2} + {\beta}^{2}} \left(\frac{\alpha}{\sqrt{{\alpha}^{2} + {\beta}^{2}}} \cos \Omega t + \frac{\beta}{\sqrt{{\alpha}^{2} + {\beta}^{2}}} \sin \Omega t\right)$

$= \sqrt{{\alpha}^{2} + {\beta}^{2}} \sin \left(\Omega t + \Phi\right)$ with $\sqrt{{\alpha}^{2} + {\beta}^{2}}$ as the amplitude.

The amplitude of the new motion is:

$A ' = \sqrt{{\left(A \sin \omega \tau\right)}^{2} + {\left(\frac{\omega A m}{\Omega M} \cos \omega \tau\right)}^{2}}$

$= A \sqrt{{\left(\sin \omega \tau\right)}^{2} + {\left(\frac{\sqrt{\frac{k}{m}} m}{\sqrt{\frac{k}{M}} M} \cos \omega \tau\right)}^{2}}$

$= A \sqrt{{\left(\sin \omega \tau\right)}^{2} + {\left(\sqrt{\frac{m}{M}} \cos \omega \tau\right)}^{2}}$

$= A \sqrt{{\sin}^{2} \omega \tau + \frac{m}{M} {\cos}^{2} \omega \tau}$

Cannot think of a simple way to verify this, so some reality checks instead:

• It is dimensionally correct

• set $m = M$ and you do get the original amplitude $A$

• set $\omega \tau = \frac{\pi}{2}$, which is the same as changing the mass when original system is at full amplitude , and $A = A '$. This is predictable. It had no momentum to preserve.

• set $\omega \tau = 2 n \pi$, which is the same as changing the mass when original system is at equilibrium , and $A ' = A \sqrt{\frac{m}{M}}$

For that last point note:

• ${\lim}_{M \to \infty} A \sqrt{\frac{m}{M}} = 0$

• ${\lim}_{M \to 0} A \sqrt{\frac{m}{M}} = \infty$.

Again there is some intuition there.