A spring has a mass of 1 kg attached to the end of it. If the spring has a spring constant of 100 N/m, then what would the frequency of this spring and mass be if it were undergoing simple harmonic motion (bouncing back and forth)? Note that this problem is asking for the frequency, "f", and not the angular frequency, W. Assume that there is no friction between the mass and the floor. O2.36 Hz O 0.71 Hz 1.15 Hz O 1.59 Hz
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 3**
**Problem Statement:**
A spring has a mass of 1 kg attached to the end of it. If the spring has a spring constant of 100 N/m, then what would the frequency of this spring and mass be if it were undergoing simple harmonic motion (bouncing back and forth)? Note that this problem is asking for the frequency, "f", and not the angular frequency, \( \omega \). Assume that there is no friction between the mass and the floor.
**Options:**
- 2.36 Hz
- 0.71 Hz
- 1.15 Hz
- 1.59 Hz
**Solution:**
To solve this problem, we use the formula for the frequency \( f \) of a mass-spring system undergoing simple harmonic motion:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
where:
- \( k \) is the spring constant (in Newtons per meter, N/m),
- \( m \) is the mass attached to the spring (in kilograms, kg),
- \( \pi \) is a constant (approximately 3.14159).
Given:
- \( k = 100 \text{ N/m} \)
- \( m = 1 \text{ kg} \)
Plug these values into the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{100}{1}} \]
\[ f = \frac{1}{2\pi} \sqrt{100} \]
\[ f = \frac{1}{2\pi} \times 10 \]
\[ f \approx \frac{10}{6.2832} \]
\[ f \approx 1.59 \text{ Hz} \]
Thus, the correct answer is:
- 1.59 Hz](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bc418ab-5051-450d-ba5a-1e906e0fb9c9%2F158386a1-54a8-43ec-a713-a0f2fb48db8a%2Fen8ktaj_processed.jpeg&w=3840&q=75)
Step by step
Solved in 2 steps with 2 images
