A vertical mass-on-a-spring system (oscillating in air) has a spring constant k

Question

A vertical mass-on-a-spring system (oscillating in air) has a spring constant k and mass m. For each of the following statements select the correct option.
1) When the length of the spring is cut in half, the spring constant a)increases b)stays the same c)decreases .
2)When the length of the spring is cut in half, the period of oscillation a)increases b)stays the same c)decreases .
3) When the mass moves from the equilibrium position to the maximum displacement, the kinetic energy of the mass a)increases b)stays the same c)decreases .
4)When the mass is doubled, the period of oscillation a)increases b)stays the same c)decreases .
5)When the mass moves from the maximum displacement to the equilibrium position, the potential energy of the spring a) increases b)stays the same c) decreases


for each one pick one of the 3 choices a)increases b) stays the same c) decreases

Explanation / Answer

A vertical mass-on-a-spring system (oscillating in air) has a spring constant k and mass m. For each of the following statements select the correct option.

1) When the length of the spring is cut in half, the spring constant a)increases b)stays the same c)decreases .

It will stay the same -- the constant is determined by the stiffness of the spring, so even if you change the length, you don't alter the spring's stiffness.

2)When the length of the spring is cut in half, the period of oscillation a)increases b)stays the same c)decreases .

T=2(m/k) --> Length of the string isn't a variable, so the period remains the same.

3) When the mass moves from the equilibrium position to the maximum displacement, the kinetic energy of the mass a)increases b)stays the same c)decreases .
KE is at a maximum when displacement is at 0, so it decreases when the spring is at maximum displacement. PE increases as the displacement increases. 4)When the mass is doubled, the period of oscillation a)increases b)stays the same c)decreases .
T=2(m/k) --> As you can see, the period will increase because mass is on top of the fraction.
5)When the mass moves from the maximum displacement to the equilibrium position, the potential energy of the spring a) increases b)stays the same c) decreases
It will decrease. I mentioned that in the KE question, but here's something more visual -- the PE of a spring=1/2kx^2, where x is the displacement. If the displacement is at the equilibrium position, it equals 0, so the PE equals 0.

for each one pick one of the 3 choices a)increases b) stays the same c) decreases

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