Forced oscillation (mass & spring)

AI Thread Summary
A 2 kg object attached to a frictionless spring is driven by an external force F = (3.00N)sin(2πt), with a spring constant of 20.0 N/m. The amplitude of the motion can be calculated using the equation A = (F0/m)/√(w² - w0²), where w is the driving frequency and w0 is the natural frequency. The correct natural frequency w0 is √(k/m), which equals 10 in this scenario. After correcting the value for w0, the amplitude calculation yields the book's answer of 5.09 cm. Proper application of these principles leads to the correct solution.
kash25
Messages
12
Reaction score
0

Homework Statement


A 2 kg object attached to a spring moves without friction and is driven by an external force given by F = (3.00N)sin(2\pit). The force constant of the spring is 20.0N/m.
Find the amplitude of the motion.


Homework Equations


I am not sure but I am trying to use:
A = (F0/m)/\sqrt{}(w^2-w<sub>0</sub>^2)^2


The Attempt at a Solution


Applying this equation using w equals 2pi and w0 is k/m (10) gives an amplitude of 0.00173. The solution in the book says 5.09cm.
Am I approaching this question correctly?
 
Physics news on Phys.org
Hi kash25,
kash25 said:

Homework Statement


A 2 kg object attached to a spring moves without friction and is driven by an external force given by F = (3.00N)sin(2\pit). The force constant of the spring is 20.0N/m.
Find the amplitude of the motion.


Homework Equations


I am not sure but I am trying to use:
A = (F0/m)/\sqrt{}(w^2-w<sub>0</sub>^2)^2


The Attempt at a Solution


Applying this equation using w equals 2pi and w0 is k/m (10)

w0 is not equal to k/m. If you correct this you should get the right answer.
 
Not exactly sure...but that equation you have might be for forced oscillations working against friction; in this case there is no friction.
 
Gear300 said:
Not exactly sure...but that equation you have might be for forced oscillations working against friction; in this case there is no friction.

No, it's the right equation. If it was a damped spring there would be another term under the radical. Using that equation (with the corrected value for w0) gives the answer given in the post.
 
remember that w0 = sqrt (k/m).

That will give you the right answer :)
 
stupid mistake..thanks for your help!
 
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