Question

A block with mass m =7.1 kg is hung from a vertical spring. When the mass...

A block with mass m =7.1 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.24 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.5 m/s. The block oscillates on the spring without friction.

1) What is the spring constant of the spring?

2) What is the oscillation frequency?

3) After t = 0.38 s what is the speed of the block?

4) What is the magnitude of the maximum acceleration of the block?

Homework Answers

Answer #1

given
m = 7.1 kg
x = 0.24 m
v_max = 4.5 m/s

1) In the equilibrium, F_spring = m*g

k*x = m*g

k = m*g/x

= 7.1*9.8/0.24

= 290 N/m <<<<<<<<<<---------------Answer

2) we know, angular frequency of motion, w = sqrt(k/m)

= sqrt(290/7.1)

= 6.391 rad/s

we know, f = w/(2*pi)

= 6.391/(2*pi)

= 1.02 Hz <<<<<<<<<<---------------Answer

3) v = vmax*cos(w*t)

= 4.5*cos(6.391*0.38) (use calculator in radian mode)

= 3.40 m/s <<<<<<<<<<---------------Answer

4) let A is the amplitude of motion.

v_max = A*w

and

a_max = A*w^2

= (A*w)*w

= v_max*w

= 4.5*6.391

= 28.8 m/s^2 <<<<<<<<<<---------------Answer

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