Simple harmonic motion and frequency; answer disagrees from answer key

In summary, a block attached to a spring oscillates in simple harmonic motion with an amplitude of 20 cm and a frequency of 2 Hz. The frequency can be found by taking the reciprocal of the period, which is 0.25 s. The answer key may be incorrect in stating that the frequency is 4 Hz.
  • #1
clairez93
114
0

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



[tex]T = 1/f[/tex]

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.
 
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  • #2
I'd say that you are correct and the answer key is wrong.
 
  • #3




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.
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which a system oscillates back and forth around an equilibrium point, with a restoring force that is directly proportional to the displacement from the equilibrium point.

2. How is frequency related to simple harmonic motion?

Frequency is the number of complete cycles or oscillations that occur in a given time period. In simple harmonic motion, the frequency is directly proportional to the square root of the force constant and inversely proportional to the mass of the system.

3. What is the formula for calculating frequency in simple harmonic motion?

The formula for frequency in simple harmonic motion is f = 1/2π √(k/m), where f is frequency, k is the force constant, and m is the mass of the system.

4. Can the frequency of simple harmonic motion be changed?

Yes, the frequency of simple harmonic motion can be changed by altering the force constant or the mass of the system. Increasing the force constant or decreasing the mass will result in an increase in frequency, and vice versa.

5. How does the amplitude of simple harmonic motion affect the frequency?

The amplitude of simple harmonic motion does not affect the frequency. The frequency only depends on the force constant and the mass of the system, not on the amplitude of the oscillations.

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