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

[Pao44445]

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?

 

[gneill]

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.

 

[Pao44445]

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

 

[gneill]

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?

 

[Pao44445]

T=2π√kM (

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?

I got it from another physics group

 

[haruspex]

View attachment 105777
I got it from another physics group

Can you not see the error in the very last step?

 

[Pao44445]

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

 

[gneill]

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.