Answer:
The maximum speed would double
Explanation:
The maximum speed of a mass-spring system is given by:
[tex]v=\omega A[/tex]
where
[tex]\omega[/tex] is the angular frequency
A is the amplitude of the motion
The angular frequency in a spring-mass system is
[tex]\omega=\sqrt{\frac{k}{m}}[/tex]
where
k is the spring constant
m is the mass
Substituting inside the first equation,
[tex]v=\sqrt{\frac{k}{m}} A[/tex]
In this problem:
- the spring constant is increased by a factor 4: k' = 4 k
- the amplitude remains constant: A' = A
So the new maximum speed would be
[tex]v'=\sqrt{\frac{4k}{m}} A= \sqrt{4} (\sqrt{\frac{k}{m}}A)= 2v[/tex]
So, the maximum speed would double.
