Simple Harmonic motion of vertical spring.

In summary, the conversation discusses the determination of the period of oscillation for an object hanging from a vertical spring. Using the equations T = 1/f, w = (k/M)^(1/2), and F=kx, the object's equilibrium at a displacement of 14cm is analyzed and the period is calculated to be 0.089s. However, a mistake in the calculation is identified and the correct answer is determined to be √70 rad/s or 11.14 Hz.
  • #1
Kizaru
45
0

Homework Statement


An object of unknown mass M is hung from a vertical spring of unknown spring constant k, and the object is observed to be at rest when the spring has extended by 14cm. The object is then given a slight push and executes SHM. Determine the period of this oscillation.


Homework Equations


T = 1/f
w = (k/M)^(1/2) = 2pi*f
F=kx

The Attempt at a Solution


The object is hanging, so it has a force of Mg pulling it down and kx pulling it up. It's in equilibrium at a displacement of 14cm (0.14m). So I set kx = Mg and solved for k:
k = Mg / x

I then plugged this into w = (k/M)^(1/2) so the M's cancel and I'm left with w = (g/x)^(1/2) = (9.8 / 0.14)^(1/2) = 70rad/s

w = 2pi*f => f = w/(2pi) = 11.14 Hz
T = 1/f = 0.089s

This isn't the right answer according to the book, and I keep getting this kind of problem wrong. I'm not entirely sure what I'm doing wrong.
 
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  • #2

The Attempt at a Solution




I then plugged this into w = (k/M)^(1/2) so the M's cancel and I'm left with w = (g/x)^(1/2) = (9.8 / 0.14)^(1/2) = 70rad/s


Check the above calculation
 
  • #3
Hi Kizaru! :smile:

(have a square-root: √ and an omega: ω :wink:)
Kizaru said:
w = (g/x)^(1/2) = (9.8 / 0.14)^(1/2) = 70rad/s

Nooo … √70 :wink:
 
  • #4
Ah yes, turns out my mistake was ignoring the square root. Thanks. I made that same mistake for all of these types of problems hehe, thanks :) Embarassing, guess I should check for elementary mistakes before I look for physics mistakes :)
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium. This means that the object will oscillate back and forth around a central point.

2. What factors affect the frequency of simple harmonic motion?

The frequency of simple harmonic motion is affected by the mass of the object, the spring constant of the spring, and the amplitude of the oscillation. A higher mass or stiffer spring will result in a lower frequency, while a larger amplitude will result in a higher frequency.

3. How do you calculate the period of simple harmonic motion?

The period of simple harmonic motion can be calculated using the formula T = 2π√(m/k), where T is the period in seconds, m is the mass of the object in kilograms, and k is the spring constant in newtons per meter.

4. What is the relationship between simple harmonic motion and energy?

In simple harmonic motion, energy is constantly being exchanged between kinetic energy and potential energy. At the equilibrium point, all energy is in the form of potential energy, while at the maximum displacement points, all energy is in the form of kinetic energy.

5. How does damping affect simple harmonic motion?

Damping, or the dissipation of energy, can affect the amplitude and frequency of simple harmonic motion. In an underdamped system, the amplitude decreases over time, while in an overdamped system, the object will eventually come to a rest. In a critically damped system, the object will return to equilibrium without any oscillation.

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