8. An object of mass 0.54 kg is attached to a horizontal spring, with spring constant 40.0 4.10 cm from the equilibrium position and released, what is its maximum speed? m/s 50 s sf60 sse €60 ssf67 If the object is stretched ssf60 ssf6cm. I ƒ60 ssf60 s³
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.
![### Problem 8: Examining Harmonic Motion
**Question:**
An object of mass 0.54 kg is attached to a horizontal spring, with a spring constant of 40.0 N/m. If the object is stretched 4.10 cm from the equilibrium position and then released, what is its maximum speed?
**Answer:**
\[ \text{_______ m/s} \]
**Explanation:**
This problem involves the principles of simple harmonic motion.
To find the maximum speed of the object, we can use the relationship between potential energy stored in the spring and the kinetic energy of the mass.
The maximum potential energy in the spring when it is stretched is given by:
\[ U = \frac{1}{2} k x^2 \]
Where:
- \( k \) is the spring constant (40.0 N/m),
- \( x \) is the displacement from the equilibrium position (4.10 cm or 0.041 m).
When the object is released and passes through the equilibrium position, all potential energy will convert to kinetic energy. The kinetic energy of the object is given by:
\[ K = \frac{1}{2} mv^2 \]
By setting the potential energy equal to the kinetic energy at maximum speed, we have:
\[ \frac{1}{2} k x^2 = \frac{1}{2} mv^2 \]
Simplifying for \( v \):
\[ v = \sqrt{\frac{k}{m}} \cdot x \]
Substitute the given values:
\[ v = \sqrt{\frac{40.0 \, \text{N/m}}{0.54 \, \text{kg}}} \cdot 0.041 \, \text{m} \]
Calculate the result to obtain the maximum speed in meters per second.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c6fb9be-cb3b-43fd-a6cc-90b4ac965f54%2Fb8fdcd87-5a30-4058-818b-b92721a0ba1d%2Fex3qln_processed.png&w=3840&q=75)
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