(a) The period will not change
The period of oscillation of a simple harmonic oscillator is given by:
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
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
[tex]E=\frac{1}{2}kA^2[/tex]
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:
[tex]E'=\frac{1}{2}k(2A)^2=4(\frac{1}{2}kA^2)4E[/tex] (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
[tex]E=\frac{1}{2}mv_{max}^2[/tex] (2)
where
m is the mass
[tex]v_{max}[/tex] is the maximum velocity
Since the total energy must be conserved, then it must be
(1) = (2)
So we can write:
[tex]\frac{1}{2}mv_{max}^2=\frac{1}{2}kA^2[/tex]
which can be rewritten as
[tex]v_{max}= A \sqrt{\frac{k}{m}}[/tex]
In this problem, the amplitude is doubled:
A' = 2A
Therefore, the new maximum velocity is
[tex]v_{max}'= (2A) \sqrt{\frac{k}{m}}=2 v_{max}[/tex]
So, the maximum velocity will double.
