A coin of mass m sits on top of a spring that oscillates vertically, up and down

Question

A coin of mass m sits on top of a spring that oscillates vertically, up and down, with small amplitude (follows Hooke's law). [m = 0.01 kg; g = 9.82 m/s2] Derive an equation that describes the motion of the coin in the vertical direction its function of time, y(t), if one assumes the coin stays in contact with the spring at all times. You may assume the motion of the spring-coin system is oscillatory. The amplitude of the oscillation is 1.2 cm. What is the maximum frequency f = omega/2pi that assures the coin remains in contact with the spring throughout?

Explanation / Answer

a) Let the spring constant be k ,l the natural lentgh of the sprimg,x the compression of the spring.The equilibrium position is given by
mg=kx...(1)
∴x=mg/k

Now the motion of the block is an shm about this mean position since the restoring force is directly proportional to the displacement of the coin from the mean position as is evident from 

F=k(x+Δx)-mg=kΔx using (1)

Hence Δx=Asinwt (A is amplitude and w is angular frequency)

y=height of equilibrium point from the ground= l-x= l-mg/k

Hence y(t)=y+Δx=l-mg/k+Asinwt (m=0.01kg,g=9.82m/s2)

b)

force on coin due to oscillation at its extreme end=mAω2=maximum force due to oscillation

force on coin due to gravitation=mg

Now,for coin to remain in contact with the spring throughout the gravitational force must be greater than the maximum possible force on the coin due to oscillation that is trying to throw it away

∴mg>mAω2

ω<√g/A

2πf<√g/A         ( f= ω/2π )

∴f<4.555 s-1

 

 

 

 

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