A 230 g block hangs from a spring with spring constant 10 N/m. At t = 0 s the bl

Question

A 230 g block hangs from a spring with spring constant 10 N/m. At t = 0 s the block is 21 cm below the equilibrium point and moving upward with a speed of 113 cm/s. What is the block's oscillation frequency? Express your answer to two significant figures and include the appropriate units. What is the block's distance from equilibrium when the speed is 60 cm/s? Express your answer to two significant figures and include the appropriate units. What is the block's distance from equilibrium at 8.0s? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

As we know that

1)

= sqrt(k/m) = sqrt(10N/m / 0.23kg) = 6.5938047362657017 rad/s frequency in radians/s
f = /2 = 1.04996 Hz

2)

At t = 0, system energy E = Ep + Ek
E = ½kx² + ½mv² = ½*10N/m*(0.21m)² + ½*0.23kg*(1.13m/s)² = 0.3673435 J
When the speed is 0.60m/s,
Ek = ½mv² = ½*0.23kg*(0.60m/s)² = 0.0414 J
So Ep = E - Ek = 0.3673435J - 0.0414J = 0.3259435J = ½kx² = ½*10N/m*x²
x = 0.255320m = 25.5320 cm

3. Well, the amplitude is found from
E = Ep = 0.3673435J = ½kx² = ½*10N/m*x²
x = 0.2710511 m = 27.10511 cm
Using the general form
x(t) = Acos(t - )
and plugging in our initial conditions:
21 = 27.10511*cos(0 - )
21/27.10511 = 0.774761653064699 = cos(-)
= - arccos0.774761653064699 = -0.684458 rad
So x(t) = 27.10511cm * cos(t + 0.684458)
When t = 8s,
x(1) = 27.10511cm * cos(6.5938047362657017*8 + 0.684458) = -26.9001291395031749 cm, max compression

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