How Do You Calculate the Period of Oscillation for a Mass on a Spring?

[Lana Elcic]

Homework Statement

A 775 g mass is hung on a spring. As a result the spring stretches 20.5 cm. If the object is then pulled an additional 3.0 cm downward and released, what is the period of the resulting oscillation?

Homework Equations

T = 2pi sqr root(m/k)
Hooke's Law Fs=kx

The Attempt at a Solution

For the period I know you use T = 2pi sqr root(m/k)

So you must find k using Hooke's Law, which we find force (m*g) and total distance is our number plus the additional 3cm, so to solve for k, its k=F/x, which you then substitute into the original equation, but I don't think my answer is right. I'm assuming my x is wrong? Do you not add the two values?

.775kg
20.5cm + 3cm = .235m

Fs=kx
(.775)(9.81) / .235m = 32.34

T = 2pi sqrroot (.775kg/32.34) = .97
97 seconds.

 

[Herman Trivilino]

How much force does it take to stretch the spring a distance of 20.5 cm?

 

[Lana Elcic]

0?

How much force does it take to stretch the spring a distance of 20.5 cm?

 

[Herman Trivilino]

How did you come up with that answer?

 

[SteamKing]

Homework Statement

A 775 g mass is hung on a spring. As a result the spring stretches 20.5 cm. If the object is then pulled an additional 3.0 cm downward and released, what is the period of the resulting oscillation?

Homework Equations

T = 2pi sqr root(m/k)
Hooke's Law Fs=kx

The Attempt at a Solution

For the period I know you use T = 2pi sqr root(m/k)

So you must find k using Hooke's Law, which we find force (m*g) and total distance is our number plus the additional 3cm, so to solve for k, its k=F/x, which you then substitute into the original equation, but I don't think my answer is right. I'm assuming my x is wrong? Do you not add the two values?

.775kg
20.5cm + 3cm = .235m

The mass was pulled down the additional 3 cm. You can't use this distance to figure out the spring constant because you don't know the total force used to stretch the spring.

Fs=kx
(.775)(9.81) / .235m = 32.34

What are the units here?

The mass by itself stretches the spring only 20.5 cm according to the problem statement.

T = 2pi sqrroot (.775kg/32.34) = .97
97 seconds.

How does .97 become 97 seconds?

 

[Lana Elcic]

How did you come up with that answer?

How did you come up with that answer?

brought to rest (lowered carefully, so Fnet = 0)

 

[Lana Elcic]

The mass was pulled down the additional 3 cm. You can't use this distance to figure out the spring constant because you don't know the total force used to stretch the spring.
So I just use the 20.5cm?

What are the units here?
(.775kg)(9.81m/s^2) / .235m = 32.34 N/m

The mass by itself stretches the spring only 20.5 cm according to the problem statement.How does .97 become 97 seconds?

I don't know

 

[Herman Trivilino]

If it takes no force to stretch a spring, that spring won't make the attached object oscillate.

 

[Lana Elcic]

If it takes no force to stretch a spring, that spring won't make the attached object oscillate.

So how do I determine the force?

 

[Lana Elcic]

I don't know

I answered the other questions within

 

[Herman Trivilino]

So how do I determine the force?

It was given in the statement of the problem.

 

[Lana Elcic]

It was given in the statement of the problem.

I'm lost.

 

[Herman Trivilino]

By definition, $k=\displaystyle \frac{F}{x}$ where $F$ is the force applied to the spring and $x$ is the distance the spring is stretched by that force.

 

[SteamKing]

So how do I determine the force?

You know that 775 grams stretches the spring 20.5 cm. You use the weight of a 775-gram mass to find the force stretching the spring.

 

[Lana Elcic]

You know that 775 grams stretches the spring 20.5 cm. You use the weight of a 775-gram mass to find the force stretching the spring.

So 9.81*.775? ...

 

[Lana Elcic]

By definition, $k=\displaystyle \frac{F}{x}$ where $F$ is the force applied to the spring and $x$ is the distance the spring is stretched by that force.

So what I would change is just use .03m instead of the .235m?

 

[Herman Trivilino]

So what I would change is just use .03m instead of the .235m?

What would be your reasoning for doing that?

Perhaps you need some time to go back and read the problem again and read through this thread again.