< Question 15 of 25 > What is the mass m of an object that is attached to a spring with a force constant of 125 N/m if 22 complete oscillations occur each 13 s? kg m =
Simple harmonic motion
Simple harmonic motion is a type of periodic motion in which an object undergoes oscillatory motion. The restoring force exerted by the object exhibiting SHM is proportional to the displacement from the equilibrium position. The force is directed towards the mean position. We see many examples of SHM around us, common ones are the motion of a pendulum, spring and vibration of strings in musical instruments, and so on.
Simple Pendulum
A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.
Oscillation
In Physics, oscillation means a repetitive motion that happens in a variation with respect to time. There is usually a central value, where the object would be at rest. Additionally, there are two or more positions between which the repetitive motion takes place. In mathematics, oscillations can also be described as vibrations. The most common examples of oscillation that is seen in daily lives include the alternating current (AC) or the motion of a moving pendulum.
![### Question 15 of 25
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**Problem Statement:**
What is the mass \( m \) of an object that is attached to a spring with a force constant of 125 N/m if 22 complete oscillations occur each 13 s?
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**Answer:**
\[ m = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]
*kg*
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**Explanation:**
To find the mass \( m \) of the object, given the force constant \( k \) of the spring and the oscillation information, we can use the formula for the period \( T \) of a simple harmonic oscillator:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
First, we need to calculate the period \( T \). Given that there are 22 complete oscillations in 13 seconds, we find:
\[ T = \frac{13 \, \text{s}}{22} \]
Next, we can solve for \( m \) by rearranging the formula:
\[ m = \frac{T^2 \cdot k}{4\pi^2} \]
By substituting the given values, we can determine the mass \( m \).
1. Calculate the period \( T \):
\[ T = \frac{13}{22} \, \text{s} \]
2. Substitute \( T \) and \( k \) into the rearranged formula:
\[ m = \frac{\left(\frac{13}{22}\right)^2 \cdot 125}{4\pi^2} \]
Finally, simplify the expression to find the mass \( m \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5039f9f1-9031-4cf0-b840-7355f91f2cd2%2F78a43763-b830-49e0-8ba5-e808ce889664%2Fgqjfuk9.png&w=3840&q=75)
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