(a) The total energy increases by a factor 4
The total energy of a simple harmonic system is given by:
[tex]E=\frac{1}{2}kA^2[/tex]
where
k is the spring constant
A is the amplitude of the motion
In this part of the problem, the amplitude is doubled:
A' = 2A
So the new total energy is
[tex]E=\frac{1}{2}k(A')^2=\frac{1}{2}k(2A)^2=4(\frac{1}{2}kA^2)=4E[/tex]
So, the energy quadruples.
(b) The maximum speed increases by a factor 2
The maximum speed in a simple harmonic motion is given by
[tex]v=\omega A[/tex]
where
[tex]\omega=\sqrt{\frac{k}{m}}[/tex] is the angular frequency, with k being the spring constant and m the mass
A is the amplitude
In this part of the problem, k and m do not change, so the angular frequency does not change. Instead, the amplitude is doubled:
A' = 2A
So the new maximum speed is
[tex]v'=\omega (A')=\omega (2A)=2 (\omega A)=2 v[/tex]
so, the maximum speed doubles.
(c) The maximum acceleration increases by a factor 2
The maximum acceleration in a simple harmonic motion is given by
[tex]a=\omega^2 A[/tex]
where
[tex]\omega=\sqrt{\frac{k}{m}}[/tex] is the angular frequency, with k being the spring constant and m the mass
A is the amplitude
In this part of the problem, k and m do not change, so the angular frequency does not change. Instead, the amplitude is doubled:
A' = 2A
So the new maximum acceleration is
[tex]a'=\omega^2 (A')=\omega^2 (2A)=2 (\omega^2 A)=2 a[/tex]
so, the maximum acceleration doubles.
(d) The period does not change
The period in a simple harmonic motion is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where m is the mass and k is the spring constant.
In this problem, the amplitude is doubled:
A' = 2A
However, we notice that the period does not depend on the amplitude, and since both m and k do not change, then the period will remain constant.
