# Resonance Frequency Homework: Find Frequency w/Given Spring & Mass

• sugz
In summary: M9 mm = AM = m" The natural frequency of the system is ω=√(k/m) which can be rearranged to find k. Then, once k is known, the frequency for resonance can be found by setting ω=√(k/m) equal to the driving frequency and solving for ω.In summary, the natural frequency of a spring-block system can be found by differentiating the displacement function twice and comparing it to Hooke's law. This results in the equation ω=√(k/m), where k is the spring constant and m is the mass of the block. To find the frequency for resonance, the
sugz

## Homework Statement

A certain spring elongates 9:0mm when it is suspended vertically and a block of mass M is hung on it. If driven by a force of variable frequency, at what frequency would resonance be obtained?a. is 0:088 rad/s
e. cannot be computed unless the value of M is given

## The Attempt at a Solution

I really don't know how to attempt this question.

If you have a driving force at various frequencies, about which frequency will give you resonance?

Isn't it when the frequency is one of the natural frequencies?

Right. And what is the natural frequency of this system?

Can't it be anything?

When you pull the mass down slightly and let it go, allowing the system to oscillate (with no driving force) at what frequency will it oscillate? This is the natural frequency of the system. Are you saying you can get it to oscillate freely at any frequency you want?

No it can't oscillate freely at any frequency but I don't know what frequency this would oscillate it in, sorry. I want to say the fundamental frequency but that is just a guess, no reasoning. Can you please explain it to me?

"The fundamental frequency" doesn't apply. (That has to do with standing waves in which there are multiple frequencies that cause a standing wave.)

Consider a function which describes the displacement of the block as a function of time. Do you know what this function might look like? (Use arbitrary constants for the unknowns.)

It would be a block in SHM, characterized by y=Acos(wt+Φ), where A is 9.0 mm?

Right now we're finding what the frequency would be if we pull the mass down and let go (the natural frequency). The amplitude would be however far we pull it down. It could be 9mm or anything else; it's irrelevant (because frequency is independent of amplitude).

Your equation is correct though. We can ignore the phase constant Φ. The phase constant Φ just shifts the "time zero"... clearly this will not effect the frequency.

Now, differentiate the function twice to get the acceleration function. Consider the time when the acceleration is maximum; what acceleration does this produce? What will be the displacement at this time? What acceleration does hooke's law produce? What must ω be in order for the accelerations to agree? Then, what is the natural frequency of system?

When deriving it, I get a= -ω^2Acos(ωt). Its a maximum when t=0. Won't the displacement at this time just be A? Acceleration by hooke's law is -kx/m =a. So ω must be the square root of the spring constant over the mass. I am sorry but I am not getting it:S

sugz said:
So ω must be the square root of the spring constant over the mass. I am sorry but I am not getting it:S
What do you mean you're not getting it... $ω=\sqrt{\frac{k}{m}}$... that's exactly right!

So then, what is the natural frequency of the object (in terms of k). And then can you find k from the information in the problem? (Find k in terms of the mass of the object.)

Some people may say it's worth memorizing ω=√(k/m) ... I say it's only worth memorizing what you unintentionally memorize.

At any rate, if you ever forget that the angular frequency ω of a mass on a spring is ω=√(k/m), then the way I showed you would be the way to derive it. Namely, differentiate the displacement function twice with respect to time, and then compare that to hookes law and solve for ω.

It's not usually worth memorizing any results, but it can sometimes be useful to 'memorize' how to derive the result.

So I'm not too sure if I understant this still but this is where I am at. According to hooke's law. -kx=ma ===> k=ma/x (1), where x is 9 mm. Acceleration is equal to
ω^2A (2). Plugging in equation 1 into ω=√ k/m i get ω=√[(ma/x)/m], which simplifies to ω=√(a/x).Substituting Equation 2 into the equation results in 1= ω=√(A/x). But I don't understand how to go beyond this point. Also, why are we only considering maximum acceleration?

sugz said:
So I'm not too sure if I understant this still but this is where I am at. According to hooke's law. -kx=ma ===> k=ma/x (1), where x is 9 mm.
If x is 9 mm then what is "a"? It can't be the acceleration of the block because the spring constant is... constant.

sugz said:
But I don't understand how to go beyond this point.
Once ω is known, how can you find the frequency? (Find the cycles per second; not the radians per second.)

sugz said:
But I don't understand how to go beyond this point. Also, why are we only considering maximum acceleration?
We only did that to derive ω2=k/m. We didn't have to choose the time when acceleration is maximum, any time would have worked.
This is how it would've gone for an arbitrary time:

Suppose the position at a particular time is ... xp=A*cos(ωT+Φ) ... where A is the position amplitude, and T is some point in time.
The acceleration at this particular time will be ... ap=-(ω2)*A*cos(ωT+Φ) ... (differentiating twice just brings out a factor of -ω2)
We can see that ap=-ω2xp
Hooke's law says map=-kxp which means that ω2=k/m

So when I said, "consider the time when the acceleration is maximum," I was just choosing a particular time T.
I could've chose any time, so I just chose the time such that cos(ωT+Φ)=1

But what is ω equal to, how do find it when we don't know k?

sugz said:
But what is ω equal to, how do find it when we don't know k?
That's the next part of the problem: we need to find k.
Look at this part of the question:
sugz said:
A certain spring elongates 9:0mm when it is suspended vertically and a block of mass M is hung on it.
Can you find k from that?

OH! I was circling around the same thing. Instead of doing mg=kx,i kept on doing ma=kx. Thank you for bearing through this with me! I finally get it!

## What is resonance frequency homework?

Resonance frequency homework involves finding the frequency of a given spring and mass system. This frequency can help determine the natural oscillations of the system and is an important concept in the study of waves and vibrations.

## Why is it important to find the resonance frequency?

Finding the resonance frequency allows us to predict the behavior of a spring and mass system. It helps us understand how the system will respond to external forces and can be used to design and optimize various mechanical and electrical systems.

## What is the equation for calculating resonance frequency?

The equation for calculating resonance frequency is f = 1/(2π√(k/m)), where f is the frequency, k is the spring constant, and m is the mass of the system. This equation is derived from the fundamental principles of oscillations and waves.

## How do you find the spring constant and mass for the equation?

The spring constant can be determined experimentally by measuring the force required to stretch or compress the spring by a certain distance. The mass can also be measured directly. Alternatively, if the system is described by a mathematical model, the values for k and m can be obtained from the model.

## What are some real-world applications of resonance frequency?

Resonance frequency is an important concept in many fields, including engineering, physics, and music. It is used in designing and tuning musical instruments, creating efficient electrical circuits, and predicting the behavior of structures such as bridges and buildings. It is also an important factor in the study of earthquakes and other natural phenomena.

• Introductory Physics Homework Help
Replies
17
Views
611
• Introductory Physics Homework Help
Replies
14
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
6K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
944