7. 215 points | Previous Answers SerPOP5 12.P.013.MI. My Notes Ask Your Teacher

Question

7. 215 points | Previous Answers SerPOP5 12.P.013.MI. My Notes Ask Your Teacher A 0.440-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.0 cm. (Assume the position of the object is at the origin at t = o.) (a) Calculate the maximum value of its speed 51.1 cm/s (b) Calculate the maximum value of its acceleration. 217.7 cm/s2 ) Calculate the value of its speed when the object is 8.00 cm from the equilibrium position. Can you determine a time when the object is at this position? You can then use this time in your equation of motion. cm/s d) Calculate the value of its acceleration when the object is 8.00 cm from the equilibrium position can/ ) Calculate the time interval required for the object to move from x = 0 to x = 4.00 cm. Need Help? 1-Read it. Master It

Explanation / Answer

From the data in the first paragraph,
= (k/m) = (8.00N/m / 0.440kg) = 4.26 rad/s
and we can model the motion as
x(t) = 12cm * sin(4.26*t)

(a) max speed = A* = 12*4.26 = 51.1 cm/s

(b) max accel = A*² = 12*4.26^2 = 217.7 cm/s²

(c) TME = ½kA² = ½ * 8.00N/m * (0.12m)² = 0.0576 J
When x = 0.08 m, PE = ½ * 8.00N/m * (0.08m)² = 0.0256 J
leaving KE = 0.032 J = ½ * 0.440kg * v²
v = 0.3814 m/s = 38.14 cm/s

(d) a = kx / m = 8.00N/m * 0.08m / 0.440kg = 1.455 m/s² = 145.5 cm/s²

(e) 4.0 cm = 12cm * sin(4.26*t)
4.26*t = arcsin(4.0/12) = 0.3398
t = 0.3398/4.26 = 0.0798 s

Hope this helps!

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