Calculating Frequency of Oscillating Mass on a Spring

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The position of a mass oscillating on a spring is described by the equation x(t) = (18.3 cm)cos[(2.35 s^-1)t]. To find the frequency of this motion, the relationship f = ω/(2π) can be applied, where ω is the angular frequency. In this case, ω is given as 2.35 s^-1, leading to the calculation of frequency as f = 2.35 s^-1 / (2π). There is some confusion regarding the relationship between frequency and period, but it is clarified that frequency is the reciprocal of the period. The discussion emphasizes the correct application of the formulas to determine the frequency accurately.
aleferesco
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Homework Statement



The position of a mass oscillating on a spring is given by x(t) = (18.3cm)cos[(2.35s^-1)t] .
What is the frequency of this motion?

Homework Equations



X= Amplitude x Cos (2pi/T x time)

The Attempt at a Solution



I know that frequency= 1/Period, I am trying to use f= 1/(2.35s^-1) but it doesn't seem correct. Also I've tried multiplying the amplitude by 2... f= 1/(18.3cm)x2


Thanks
 
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so the form is x(t) = Acos( \omega t )

where \omega = 2 \pi f

Can you see how this helps you?
 
so omega= 2.35s^-1

and so to find frequency I could just do f=2.35s^-1/2pi

I'm not sure about the reversing 1/1/T = s^-1
 
aleferesco said:
[...]

I'm not sure about the reversing 1/1/T = s^-1

Sorry I don't understand what you're saying here, but yes:

f = \frac{\omega}{2\pi}

are you confused because omega is a reciprocal? That should play no part in exchanging frequency with period. So just follow that formula and you should be fine.

Note: I have not solved the problem numerically, but I can tell it's going to be a very small answer.
 
Thank you very much!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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