(a) The period will not change
The period of oscillation of a simple harmonic oscillator is given by:
![T=2\pi \sqrt{\frac{m}{k}}](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%7D)
where
m is the mass
k is the spring constant
As we can see from the equation, the period of oscillation does not depend on the amplitude: therefore, if the amplitude of the oscillator is doubled, the period will not change.
(b) The total energy will quadruple
The total mechanical energy of a simple harmonic oscillator is given by
![E=\frac{1}{2}kA^2](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7DkA%5E2)
where this term represents the maximum elastic potential energy when the spring is completely compressed/stretched (so, when kinetic energy is zero), and where
k is the spring constant
A is the amplitude
In this problem, the amplitude is doubled:
A' = 2A
Therefore, the new total energy will be:
(1)
So, the total energy will quadruple.
(c) The maximum velocity will double
The maximum velocity of the mass oscillating is achieved when the mass crosses the equilibrium position: at that point, the elastic potential energy is zero (because the displacement is zero), and so the total energy is simply
(2)
where
m is the mass
is the maximum velocity
Since the total energy must be conserved, then it must be
(1) = (2)
So we can write:
![\frac{1}{2}mv_{max}^2=\frac{1}{2}kA^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv_%7Bmax%7D%5E2%3D%5Cfrac%7B1%7D%7B2%7DkA%5E2)
which can be rewritten as
![v_{max}= A \sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=v_%7Bmax%7D%3D%20A%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
In this problem, the amplitude is doubled:
A' = 2A
Therefore, the new maximum velocity is
![v_{max}'= (2A) \sqrt{\frac{k}{m}}=2 v_{max}](https://tex.z-dn.net/?f=v_%7Bmax%7D%27%3D%20%282A%29%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D%3D2%20v_%7Bmax%7D)
So, the maximum velocity will double.