Springs and simple harmonic motion

AI Thread Summary
A mass oscillating on a spring with a period of 1.45 seconds has an initial position of 8.80 cm and zero speed at t = 0. The angular velocity is calculated as 4.33 rad/s, leading to an amplitude of 8.80 cm. The user attempts to find the acceleration at t = 2.30 seconds using the formula for acceleration, but initially arrives at an incorrect value of 142 m/s². The error was identified as a failure to convert the amplitude from centimeters to meters before calculation. Correcting this conversion resolves the confusion and leads to the accurate acceleration value.
Jm4872
Messages
12
Reaction score
0

Homework Statement



A mass is oscillating on a spring with a period of 1.45 s. At t = 0 the mass has zero speed and is at x = 8.80 cm. What is the magnitude of the acceleration at t = 2.30 s?

Homework Equations


x(t)=Acos(wt + phi)
a(t)=-Aw^2cos(wt + phi)
a=w^2x
T=2pi/w


The Attempt at a Solution



I've tried this a few different ways, first off what I do is convert the period (1.45s) to angular velocity using T=2pi/w, this then gives me the result of 4.33 rad/s. Using this I then substitute values into the equation for position to get,
8.80 = Acos(0)
from this equation I then solve for A and receive 8.80 cm.
Now the information I have is..
w=4.33 rad/s
A=8.80cm
t=2.30s
All of this I put into my equation for acceleration to get..
a(2.30)=-8.80(4.33^2)cos(4.33*2.30)
solving for acceleration the result comes to be 142 m/s^2, however when I input this number it comes out to be wrong? I'm confused as to what it is that I'm doing wrong, please help!
 
Physics news on Phys.org
You just forgot to convert the amplitude of the oscillation (which is in cm) to metres before computing your answer.
 
oh wow haha now I feel dumb!
haha thank you!
 
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