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.
I would assume the amplitude would decrease if mass was added, but I don't know how to prove it.
1 Answer
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, qquad omega = sqrt(k/m)# 
#dot x = omega A cos omega t#
At displacement
#m dot x = M dot xi# , where#dot xi# is the new velocity#= m/M dot x(tau)#
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.
Which will solve as:

#xi = P cos Omega t + Q sin Omega t, qquad Omega = sqrt(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 = (omega A m)/M cos omega tau = Omega Q implies Q = (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 + (omega A m)/(Omega M) cos omega tau sin Omega t #
To find the new amplitude, rewrite as:

#xi = alpha cos Omega t + beta sin Omega t # 
#xi =sqrt(alpha^2 + beta^2) ( alpha/sqrt(alpha^2 + beta^2) cos Omega t + beta/sqrt(alpha^2 + beta^2) sin Omega t ) #
The amplitude of the new motion is:
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 = 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(m/M)#
For that last point note:

#lim_(M to oo) A sqrt(m/M) = 0# 
#lim_(M to 0) A sqrt(m/M) = oo# .
Again there is some intuition there.