Simple harmonic motion and frequency; answer disagrees from answer key

[clairez93]

Homework Statement

A block attached to a spring oscillates in simple harmonic motion along the x axis. The limits of its motion are x = 10 cm and x = 50 cm and it goes from one of these extremes to the other in 0.25 s. Its amplitude and frequency are:

A) 40 cm, 2 Hz
B) 20 cm, 4 Hz
C) 40 cm, 2 Hz
D) 25 cm, 4 Hz
E) 20 cm, 2 Hz

Homework Equations

$$T = 1/f$$

The Attempt at a Solution

I got the amplitude correct by adding the limits and dividing by two to find the midpoint of the motion, which was 30 cm, which I took to be equilibrium. Then I subtracted 10 cm, one limit of motion, from 30 cm, equilibrium to get the amplitude of 20 cm.

The frequency is where I mess up. I multiplied .25s by 2, since I thought it takes .25 seconds to complete half the cycle; I thought it must go back to the same extreme to complete a cycle, and thus would take another .25s. I got .5s for the period and then took the reciprocal of .5 for a frequency of 2 Hz.

The answer key however says the answer is B, a frequency of 4 Hz. Is this a mistake of the answer key or on my part? I do not see why the frequency would be 4.

 

[Doc Al]

I'd say that you are correct and the answer key is wrong.

 

[thomate1]

It is possible that there is a mistake in the answer key. Your reasoning for calculating the frequency is correct and should result in a frequency of 2 Hz. It is also important to note that the frequency of a simple harmonic motion is determined by the spring constant and the mass of the block attached to the spring. Therefore, it is possible that the answer key is incorrect or that there is missing information in the problem that would change the frequency. It would be helpful to have more context or information in order to determine the correct answer.