Calculating Frequency of Oscillating Mass on a Spring

AI Thread Summary
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
Messages
26
Reaction score
0

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
 
Physics news on Phys.org
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!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top