Question

A 4.70 kg block hangs from a spring with spring constant 2300 N/m . The block is pulled down 5.50 cm from the equilibrium position and given an initial velocity of 1.10 m/s back toward equilibrium.

1) What is the frequency of the motion?

2) What is the amplitude?

3) What is the total mechanical energy of the motion?

Answer 1

1 )for frequency use k = F/R = m*w^2,where w is the angular frequency in radians per second.
and k is the spring constant.
To convert w to frequency remember frequency = w/(2*π)

So, 2300 = 4.7 * w^2

w = 22.12 rad/s

now frequency,

f = w/(2*π)

f = 22.12/(2*3.14)

f = 3.52 Hz

so, the frequency of the oscillation is 3.52 Hz.

3 )When the block is pulled down 5.5 cm the system gains a net energy = (1/2) * k *x^2

= 1/2 * 2300 * 0.055^2= 3.48 J

And the kinetic energy given by that initial velocity,

=(1/2)* m * v^2 = (1/2) * 4.7 * 1.1^2

in kinetic. = 2.84 J

Add them up to get the total mechanical energy in the oscillation. =  

3.48 + 2.84 = 6.32 J

2) For the amplitude, at the maximum extension then (1/2) * k * x^2 = 6.32 J (total energy)

so, 0.5 * 2300 * x^2= 6.32

x^2 = 0.00549 m²

x = 7.4 cm

thus, the amplitude of the oscillation is 7.4 cm.

final answers :-

1) 3.52 Hz

2) 7.4 cm

3) 6.32 J

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