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Simple Harmonic Motion of mass is suspended by a spring

  • Thread starter Magma828
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  • #1
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A mass is suspended by a spring. The mass is pulled down and released at t=0. The equation for its displacement x in m from the equilibrium position is x = 0.050 cos(πt).

π = pi = 3.14159 for clarification (because it looks like an n for some reason)

(a) What is (i) the amplitude, (ii) the frequency and (iii) the period of the oscillation?
(b) What is (i) the displacement and (ii) the acceleration of the mass at t = 0.50s, 0.75s, 1.5s, 1.8s?
(c) What is the velocity of the mass at t = 0.50s, 1.0s?

You may also need x = A cos(2πft) where x = displacement, A = amplitude, f = frequency and t = time period.

I'm not asking for someone to do the entire question, I'm just stuck with certain parts and I'd be able to do the rest if I knew how to do earlier bits. I've posted the entire question for clarity.

The answer to part (a)(i) is 0.05m which I THINK is given by x = A = 0.050 cos(π0). I cannot find the frequency in (a)(ii), and posting all the crazy workings I've tried would probably confuse you. The period in (a)(iii) would simply be t=1/f.

I also cannot do any of part (b)(i) where it wants the displacement of the mass after time t. I have not yet attempted (b)(ii) and (c) because I probably need previously worked out data.
 

Answers and Replies

  • #2
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A mass is suspended by a spring. The mass is pulled down and released at t=0. The equation for its displacement x in m from the equilibrium position is x = 0.050 cos(πt).

π = pi = 3.14159 for clarification (because it looks like an n for some reason)

(a) What is (i) the amplitude, (ii) the frequency and (iii) the period of the oscillation?
(b) What is (i) the displacement and (ii) the acceleration of the mass at t = 0.50s, 0.75s, 1.5s, 1.8s?
(c) What is the velocity of the mass at t = 0.50s, 1.0s?

You may also need x = A cos(2πft) where x = displacement, A = amplitude, f = frequency and t = time period.

I'm not asking for someone to do the entire question, I'm just stuck with certain parts and I'd be able to do the rest if I knew how to do earlier bits. I've posted the entire question for clarity.

The answer to part (a)(i) is 0.05m which I THINK is given by x = A = 0.050 cos(π0). I cannot find the frequency in (a)(ii), and posting all the crazy workings I've tried would probably confuse you. The period in (a)(iii) would simply be t=1/f.

I also cannot do any of part (b)(i) where it wants the displacement of the mass after time t. I have not yet attempted (b)(ii) and (c) because I probably need previously worked out data.
for part (a) ii, I presume that you've studied SHM in class, now you know that the "proper" displacement equation for SHM is [tex] x = Acos(2\pi ft)[/tex], indeed, it's given in the question, so in this example since you know that [tex] 2\pi f = \pi [/tex], not sure how solid i am there, but see if it works out (do you have the answers?)

for (b) i) it wants you to find x given certain times, since you have the equation for displacement at the top i sugguest you plug in the values given to you for t to work out x.

hint for (b) ii), since you know an equation for x, displacement, do you know how to find an equation for velocity and therefore acceleration?
 
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  • #3
Delphi51
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For aii, compare x = A cos(2πft) with x = 0.050 cos(πt).
Just a matter of setting 2πft = πt and solving for f.

For bi, evaluate x = 0.050 cos(πt) at the given time. x is the displacement.
To find the velocity use v = dx/dt, just differentiate the x = 0.050 cos(πt) formula. Acceleration = dv/dx.
 
  • #4
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Thanks for the responses guys, much appreciated. I'm having trouble understand what you're saying though.

Chewy0087, what do you mean by 2πf=π? Is that a typo?

Delphi51, you just completely screwed me over with the terminology :P
What do you mean by comparing x = A cos(2πft) with x = 0.050 cos(πt)? For the first, I don't know f or t and for the second I don't know t.

If c involves differentiation I wont be able to do it because I don't do A-Level Maths and haven't done calculus. That's not an issue anyway, I just want to do the basic stuff first.

The methods you both suggested for bi sound like what I did, but I'll concentrate on working out that correctly when I've found the frequency.

The answer in the text book for frequency is 0.50Hz.
 
  • #5
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Chewy0087, what do you mean by 2πf=π? Is that a typo?
No this is the whole point, they tell you at the start that for this system;

[tex] x = 0.05cos(\pi t) [/tex]

but you're also told that

[tex] x = 0.05cos(2\pi ft) [/tex] or for any system as a general rule - [tex] x = Acos(2\pi ft) [/tex]

so if you equate them, [tex] 2 \pi ft = \pi t [/tex]

therefore; [tex] 2f = 1 [/tex]

do you see what i did there?
 
  • #6
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Oh yeah I get it now, so you're saying that the frequency is always 0.5Hz in SHM?
 
  • #7
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No no no!

it's only in this example, the only reason why it's 0.5Hz is because they tell you that [tex] x = 0.05cos(\pi t) [/tex] this would be different depending on the scenario etc

for acceleration the general rule is like so [tex] a = -(2\pi f)^2 x[/tex]
 
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  • #8
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Okay thanks, I've managed to do up to part c now (not including). I can't differentiate because I've never done calculus, can anyone explain another method?
 
  • #9
Delphi51
Homework Helper
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You can probably look up the general formula for velocity, similar to the one you have for displacement:
x = A cos(2πft)
v = -2πfA sin(2πft)
acceleration -4π²f²A cos(2πft)
It is very surprising that you were asked for velocity and acceleration when you haven't done calculus, though. Maybe I'm missing something.
 
  • #10
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It mentions calculus very briefly in this section of the book, but says we don't actually have to know it. Most people who do Physics also do Maths anyway, but I don't.. :(

v = -2πfA sin(2πft) worked though woo :) This equation is listed in the book as v=Aw.cos(wt) where w=2πf which is the same as the one you gave except cos instead of sin.. I was wondering why it wouldn't give me the right answer.

Is it always sin for velocity, or should I just use the opposite of the function used for displacement?
 
  • #11
370
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No no no!

it's only in this example, the only reason why it's 0.5Hz is because they tell you that [tex] x = 0.05cos(\pi t) [/tex] this would be different depending on the scenario etc

for acceleration the general rule is like so [tex] a = -(2\pi f)^2 x[/tex]

the general formulae for acceleration is

[tex] a = -(2\pi f)^2 x[/tex] i'm afraid you can't derivate it yourself unless you've studied calculus..., x of course is displacement, another way of writing it is;

[tex] a = -(2\pi f)^2 Acos(2\pi f t)[/tex]

and yes in SHM sin is velocity, this is part of calculus but when you "differentiate" displacement, x, you get v, velocity, so when you differentiate cos(x) you get -sin (x).

you'll study this in much greater detail soon, if you don't mind me asking what country/grade are you in? =D
 

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