Question

A block with mass m =7.3 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.29 m. While at this equilibrium position, the mass is then given an initial push downward at v = 5 m/s. The block oscillates on the spring without friction.

1) What is the spring constant of the spring? N/m Submit

2) What is the oscillation frequency? Hz Submit

3) After t = 0.45 s what is the speed of the block? m/s Submit

4) What is the magnitude of the maximum acceleration of the block? m/s2 Submit

5) At t = 0.45 s what is the magnitude of the net force on the block? N Submit

6) Where is the potential energy of the system the greatest? At the highest point of the oscillation. At the new equilibrium position of the oscillation. At the lowest point of the oscillation. Submit

Answer #1

1) Mass of block = m=7.3 kg

weight of block =W= 7.3*9.8=71.54 N

length by which spring stretches = 0.29 m

spring constant =k= weight/stretch length = 71.54/0.29=246.7 N/m

2) oscillation frquency is given by f = (1/2*pi)*sqrt(k/m)=(1/2*pi)*sqrt(246.7/7.3)=0.925 Hz

3)the equation of motion is given by x(t)=Asint

where A is amplitude of vibration, =angular velocity,x(t) is displacement in time t

=2*pi*f=2*pi*0.925=5.81 rad/s

velocity is given by v=dx/dt = Acost

given , at t=0,v=5 m/s

or, Acos (0)=5

Or, A=5

A=5/=5/5.81=0.86m

speed of block after 0.45 seconds= Acos(*0.45)=5*cos(5.81*0.45)=-4.32 m/s(negatiove sign means velocity is downwards)

4)acceleration is given by d^{2}x/dt^{2} =
-A^{2}sint

maximum acceleration = A^{2}=0.86*5.81^{2}=29.03
m/s^{2}

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