A 240 g block hangs from a spring with spring constant 13 N/m . At t=0s the bloc

Question

A 240 g block hangs from a spring with spring constant 13 N/m . At t=0s the block is 28 cm below the equilibrium point and moving upward with a speed of 113 cm/sA 240 g block hangs from a spring with spring constant 13 N/m . At t=0s the block is 28 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.

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Correct

Part B

What is the block's distance from equilibrium when the speed is 72 cm/s ?

Express your answer to two significant figures and include the appropriate units.

0.263m

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Part C

What is the block's distance from equilibrium at t=8.0s?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

A) = sqrt(k/m) = sqrt(13N/m / 0.24kg) = 7.36 rad/s frequency in radians/s
f = /2 = 1.17 Hz frequency in Hz, which is what they want

B) At t = 0, system energy E = Ep + Ek
E = ½kx² + ½mv² = ½*13N/m*(0.28m)² + ½*0.24kg*(1.13m/s)² = 0.663 J
When the speed is 0.72 m/s,
Ek = ½mv² = ½*0.24kg*(0.72m/s)² = 0.062 J
So Ep = E - Ek = 0.663J - 0.062J = 0.601J = ½kx² = ½*13N/m*x²
x = 0.304 m = 30.4 cm

C) Well, the amplitude is found from
E = Ep = 0.663J = ½kx² = ½*13N/m*x²
x = 0.32 m = 32 cm
Using the general form
x(t) = Acos(t - )
and plugging in our initial conditions:
28 = 32*cos(0 - )
28/32 = 0.875 = cos(-)
= - arccos0.773 = -0.505 rad
So x(t) = 32cm * cos(t + 0.505)
When t = 8 s,
x(8) = 32cm * cos(7.36*8 + 0.505) = -30.52 cm

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