# Spring Mass SHM problem

allykat

## Homework Statement

A mass attached to a horizontal spring is pulled to the right at t=0. The mass passes x=+10 cm at t=0.3s with a velocity 85.7cm/s. What is a) frequency b)amplitude c) phase constant

## Homework Equations

Acos(ωt+phi)= x(t)
-ωAsin(ωt+phi)=v(t)

## The Attempt at a Solution

knowing that Acos(ω0.3+phi)= 10
-ωAsin(ω0.3+phi)=85.7cm/s
diving v(t) by x(t) gets
-ωtan(ωt+phi)=85.7/10=8.57s but
and for phi since at t=0 x I assume is A but I am not sure so 1=cos(0+phi) so phi would be zero
and amplitude is hard to find but it is the original stretched amount I think.

Basic_Physics
Correct phi = 0 rad so x = A cos(ωt) and v = -Aωsin(ωt)

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Homework Helper
phase is zero for the reason you said - well done.
you forgot the value of t in the equation with the tan in it.
also pay careful attention to the sign of the velocity ...
- if the mass is pulled to the right, is it's x position positive or negative?
- therefore, just after it is released, which direction is it travelling - i.e. what is the sign of the velocity?
- which direction is it travelling when it "passes x=+10cm"?

It can help if you sketch out the cosine curve.
You also know how energy conservation applies to this system.

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Mentor
The problem seems terribly underspecified to me. It forces one to make all sorts of assumptions which may or may not be correct, not the least of which is the assumption that the reference frame has +x to the right and the equilibrium point f the spring as the origin.

"Pulled to the right" is pretty vague. It could mean pulled to an unspecified distance and held stationary before release, or it could mean pulled and given an initial velocity at the origin, or it could mean pulled to some unspecified distance and given some unspecified velocity.

Suppose the mass is pulled to the right at t = 0 and released with an initial positive velocity at the equilibrium point. Then the mass could be passing +10 cm on the first outward journey. Or the mass could be released from rest some distance past the 10 cm point, and subsequently pass the +10 cm point outward bound after having gone through a cycle (so it's the second time past the point, the first inward-bound trip having a negative velocity). There are several such scenarios which could influence the results.

allykat
Unfortunately that's how underspecified the problem is in the textbook which makes it hard for me to answer the question.

Mentor
Unfortunately that's how underspecified the problem is in the textbook which makes it hard for me to answer the question.

Was the problem accompanied by a diagram?

nasu
Suppose the mass is pulled to the right at t = 0 and released with an initial positive velocity at the equilibrium point.
It's quite common to associate "released" with zero initial velocity. Same as for free fall motion. A ball released from height h is understood to have zero initial velocity.

So the initial condition I think it is specified quite well.
The signs are not, true. But this has no consequence on the quantities to be found.

Mentor
It's quite common to associate "released" with zero initial velocity. Same as for free fall motion. A ball released from height h is understood to have zero initial velocity.

So the initial condition I think it is specified quite well.
The signs are not, true. But this has no consequence on the quantities to be found.

I strongly disagree. You're attributing MY words to the problem now. The problem statement did not even use the word "release"! Nothing is stated about how the mass was "pulled" or how it was set in motion. And assuming that there are sign errors in given values is pretty far down my list of "reasonable interpretation" coping mechanisms...

All this to say that one's personal biases for what constitutes reasonable assumptions comes into play (and I'll bet that mine are no "better" than yours!).

nasu
You are right, sorry.
I was commenting on your use of the world "release" and I assumed it is from the problem.
And considered this a "reasonable assumption".

The text of the problem is indeed a little cryptic.

Homework Helper
Yep "pulled to the right" could just mean "given a yank" ... but lamenting over how badly phrased the problem statement is doesn't get us anywhere.

When approaching poorly specified problems in texts, as in real life, you need to draw from the metadata - the context. The problem statement needs to be read in the context of the general style of the text book - which we don't have. Allykat will have to make that particular judgement call - it's part of the exercise anyway.

* x positive to the right would be Cartesian - a pretty common default in texts.
* the author could have intended us to understand "pulled to the right" to mean "given an initial velocity to the right" ... that would actually jell well with the subsequent information. I think the alternate interpretation: "displaced to the right and released" is supported by the way allykat automatically reached for the cosine function - suggesting earlier context.

Given the instruction "pull the mass to the right to start it oscillating" - which would you be naturally inclined to do?

This is a judgement call that has to be made in context that we don't have.

In the absence of other information, I'd investigate both situations.
Assumptions must be clearly stated at the start of the working.

There would be infinite possible solutions even so, due to aliasing.
A useful approach is to work out the soonest that the conditions could occur - if you wanted, you could include the other possibilities as a function of an integer n, but the lowest frequency solution is probably the one intended.

I'd take the book at it's word, though, and notice that a positive velocity passing through a positive displacement will restrict the situation in a particular way, which is different depending on how you interpret "pulled to the right".

Either way it looks like it will be fairly involved to work out - unless there is some preformatted solution already in the course notes or earlier in the text.

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allykat
There was no diagram and this is a new book that our school uses. I am quite frustrated by the lack of info as well.

Homework Helper
There was no diagram and this is a new book that our school uses. I am quite frustrated by the lack of info as well.
That's all right ...

From your assumptions, you managed to get an equation with only stuff you know and the frequency. Have you worked out how to solve it yet?

Have you attempted the questions posed to guide your thinking in post #3?

Have you gone through the chapter on SHM again to see if there is some physics in there to help you?

Have you tried sketching the x(t) and v(t) curves for a few periods to see what that tells yu about the solutions?

Has this problem been set as homework so there is someone to ask to clarify the problem?

Which education level is the text for. For that matter which text is it (title, author, publication date please) - one of us may have it.

allykat
It's for first year university homework and it's a custom text they made this year. It's full of a lot of errors as the professors pointed out. Also one of our professors wrote it.

Homework Helper
OK then - ask other students and, if the others are also mystified, ask the prof as a group.
It may be the open-ended nature is intended though - part of the exercise is to make these decisions.
Even if it is not - you will be showing "initiative" and "promise" worth grades and special notice if you tackle it anyway.

You don't seem to be interested in doing the problem though.
You have the option of just throwing up your hands and surrendering - that's a valid response... or you can get stuck in and do the problem. Up to you. Let me know when you are ready.

allykat
Hello, it seems you have misunderstood, I have been doing this problem for 3 days now looking at various resources, but the professor being really busy this week, I wondered that the physics forum may lend me some aid. Unfortunately, it has been assumed that I don't really care about a problem that I'm still working on. I know about other students and other such resources but sometimes those resources do not work out. I think I let you know I care about this problem when I first posted it.

Mentor
Hi allykat. If this is an assignment that is to be handed in for marking by a person, then a valid approach is to state why the problem is not well posed and show that it can be interpreted in different ways. Then say that you will choose a likely interpretation, clearly state it, and solve that.

You will be demonstrating the required knowledge of the subject by solving a valid interpretation of the problem.

For extra points you can solve more than one scenario of the problem

Homework Helper
Hello, it seems you have misunderstood, I have been doing this problem for 3 days now looking at various resources, ... , it has been assumed that I don't really care about a problem that I'm still working on.
Glad to hear you do want to do the problem - and, hopefully, do it yourself.
I look forward to hearing your answers to the questions that have been put to you.
Those questions are not rhetorical - they are supposed to help you think about the problem more clearly and to help us see what you are doing and how you think.

So far, I cannot tell what you are doing. You have to show me.

allykat
First off, these are practice problems not to be handed in. We're given the answers to check up on ourselves. It says in the aswers that the frequency is 7.7 rad/s and the amplitude is 15cm. I keep diving v(t) by x(t) to get -ωtan(ωt)=85.7/10=8.57s but I don't know how to proceed after this step. I know the mass is going to the left hence the amplitude is greater than 10 and the mass is moving towards a negative displacement.

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Homework Helper
You say you got that equation by dividing v(t) by x(t)?
So what happened to the v(t)?
... OH I see, you subbed the numbers in ...

... this amounts to finding the intersection between the curve ##y_1(\omega)=x(0.3)/\omega## and ##y_2(\omega)=v(0.3)\tan\big((0.3)\omega\big)## ... which you may prefer to do numerically - i.e. via Newton-Raphson. A representative sketch of v(t) and x(t) will show you what to pick as a first guess.

If you plot y1 and y2 on the same axis, you'll see there are an infinite number of discrete solutions.
Maybe one will match.

Is there some reason you won't answer questions?
If I restrict myself to answering your questions as they are posted I will only reinforce bad ideas.

allykat
... this amounts to finding the intersection between the curve ##y_1(\omega)=x(0.3)/\omega## and ##y_2(\omega)=v(0.3)\tan\big((0.3)\omega\big)## ... which you may prefer to do numerically - i.e. via Newton-Raphson. [/QUOTE said:
I don't understand what the above equations are? (a little confused on the wording). What I know is that for frequency I would have to find omega but the amplitude is not given. I am not asking for the answer (since I know it) but specifically how I get rid of tan(ωt) when I have -ωtan(ωt)=85.7/10=8.57s since taking the tan of both sides gets me stuck with tan (-ω)(ωt)= tan 8.57s and again I have tan of the omega which I am trying to find, is there a specific mathematical identity I'm missing?

Homework Helper
You also have an equation which does not have the amplitude in it.
It's one equation with one unknown - you can solve it.

On notation:
x(0.3) means the value that x takes at t=0.3.
When you have a relation f(x)=g(x) it is asking for the intersection of f(x) with g(x).
f and g are two curves - if the solution exists, then they will intersect at one or more points.

Lets recap:

At ##t=0##, ##x(0)=x_0##, ##v(0)=v_0## ... which you are not given, so this is just notation.

At some time later: ##t=a##, ##x(a)=x_a##, ##v(a)=V_a## ... which you are given. a=0.3 etc right?

If the mass is displaced by a distance A and ten released, then: ##x_0=A## and ##v_0=0##
And you have equations of motion:

(1) ##x(t)=A\cos\omega t## and (2) ##v(t)=-\omega A\sin \omega t##

(2)/(1) gives you:

##v(t)=-\omega x(t)\tan\omega t##

plug in your known values and divide through by the frequency:

$$\frac{v_a}{\omega} = -x_a\tan\omega t$$ (BTW: did you misplace a minus sign?)

This should be where you are up to - I've rearranged things a bit. Let me know if I've lost you.
Note: you cannot get rid of the tan function.

Anyway: this is an equation of the form ##f(\omega)=g(\omega)## ... i.e. the solutions are the values of ##\omega## where ##f(\omega)## intersects ##g(\omega)##.

Use your computer to draw you a graph of ##f## and ##g## vs ##\omega## and where those lines intersect are your solutions.

I'll start you off:

... above: the horizontal axis is frequency, the green line is f and the blue line is g
I've marked the solutions in with a red x.
Notice how the periodicity of the tangent function in g means there are an infinite number of solutions. Your prof seems to intend you to pick the first one ... you should be able to see the frequency is about 7rad/s ... you need more accuracy than that :)

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allykat
I graphed it and got the answer for omega is 7.7 rad/s. I was wondering if there is a way to do this algebraically as we won't be allowed graphing calculators on our exams. Btw thank you for helping me with this, it makes more sense now.