Time Period of SHM | 500kg Car w/ 196,000 N/m

In summary, the conversation discusses finding the time period of a car in simple harmonic motion after the driver gets out. The given data includes the mass of the car and the spring constant, and the equation for the time period is T = 2π√(M/k). However, there was a mistake in the final step of the derivation in the given equation, resulting in the argument under the square root being the reciprocal of what it should be. The correct equation is T = 2π√(k/M) and when using this equation, a time period of 0.317 seconds was obtained.
  • #1
Pao44445
47
0

Homework Statement



a man sits in a car that makes the center gravity of the car is pulled down by 0.3 cm. After he gets out of the car, find the time period of the car while it is moving in SHM

Mass of the car = 500kg
Spring constant = 196,000 N/m

Homework Equations


?

The Attempt at a Solution


I know that mass and the spring constant, both affect the time period of the object but how is their equation?
 
Last edited:
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  • #2
The mass and spring constant are given data, not relevant equations.

Check your course text or notes or simply search the web for the period of a mass-spring oscillator.
 
  • #3
gneill said:
The mass and spring constant are given data, not relevant equations.

Check your course text or notes or simply search the web for the period of a mass-spring oscillator.
sorry, I've edited it.
I found this equation
T=2π(SQRT*K/M) but I don't know if it correct or not. I plugged the the spring constant and the mass in the equation, I got about 124.33 s, I think it is too high for the time period
 
  • #4
Presumably you meant:
$$T = 2 \pi \sqrt{\frac{k}{M}}~~~~~~~~\text{(which is not correct)}$$
You should be able to check an equation to see if the units make sense. Remember that a Newton (N) can be broken down as ##N = kg~m~s^{-2}## (which should be obvious from Newton's second law, f = ma).

In this case I think you'll find that the k and M should switch positions. Where did you find your version of the equation?
 
  • #5
gneill said:
T=2π√kM (
gneill said:
Presumably you meant:
$$T = 2 \pi \sqrt{\frac{k}{M}}~~~~~~~~\text{(which is not correct)}$$
You should be able to check an equation to see if the units make sense. Remember that a Newton (N) can be broken down as ##N = kg~m~s^{-2}## (which should be obvious from Newton's second law, f = ma).

In this case I think you'll find that the k and M should switch positions. Where did you find your version of the equation?
14322768_893995810745305_4438087625792457269_n.jpg

I got it from another physics group
 
  • #7
haruspex said:
Can you not see the error in the very last step?
When he converted ω is equal to 2π/T?
then it should be

ω = (SQRT*k/M)
2π/T = (SQRT*k/M)
T = 2π / (SQRT*k/M)

am I right? I used this equation an I got about 0.317 s
 
  • #8
That works. The answer looks good.

The formula is usually written as ##T = 2 \pi \sqrt{\frac{M}{k}}##.

In the scanned page that you attached in post #5 there was an error in the final step of the derivation, as pointed out by @haruspex. It resulted in the argument under the square root being the reciprocal of what it should.
 

1. What is SHM (Simple Harmonic Motion)?

SHM is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction of the displacement. It is characterized by a sinusoidal pattern and is observed in many physical systems, including springs, pendulums, and vibrating objects.

2. How is the time period of SHM calculated?

The time period of SHM can be calculated using the formula T = 2π√(m/k), where T is the time period, m is the mass of the object, and k is the spring constant. In the case of a 500kg car with a spring constant of 196,000 N/m, the time period would be T = 2π√(500/196,000) ≈ 0.142 seconds.

3. Why is the time period of SHM independent of amplitude?

The time period of SHM is independent of amplitude because the restoring force and the displacement are directly proportional. This means that as the amplitude increases, the force also increases, resulting in the same time period. This principle is known as Hooke's Law.

4. What factors can affect the time period of SHM?

The time period of SHM can be affected by the mass of the object, the spring constant, and the amplitude of the motion. In addition, factors such as friction and air resistance can also play a role in altering the time period.

5. How can SHM be applied in real-life situations?

SHM has many practical applications, including in the design of springs for various purposes such as shock absorbers in cars and mattresses. It is also used in the study of earthquake vibrations, pendulum clocks, and musical instruments such as guitars and violins.

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