When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled down from this position and released. It undergoes 53 oscillations in 30.6 s. What was the distance d?

The answer is supposed to be 8.28 cm. How can I get this answer? Thank you in advance.

One of the related questions helped me. Thank you Jiskha!

To find the distance d, we can use the formula for the period of oscillation of a mass-spring system.

The period (T) of an oscillating mass attached to a spring is the time it takes for one complete cycle of oscillation. In this case, we are given that the mass undergoes 53 oscillations in 30.6 s.

First, we need to calculate the time it takes for one oscillation:

T = 30.6 s / 53 oscillations

Next, we can use the formula for the period of a mass-spring system to find the distance d:

T = 2π √(m / k)

where m is the mass of the object and k is the spring constant. In this case, we are given the period T and we need to solve for d.

Rearranging the formula, we have:

d = T^2 * k / (4π^2 * m)

To find d, we need to know the mass (m) and the spring constant (k). Assuming these values are not given, we cannot calculate the distance d with the information provided unless we have additional data.

If you have access to the mass (m) and the spring constant (k), you can substitute these values into the formula above to find the distance d.