A bolt of mass 1.90×102 kg moves with SHM that has an amplitude of 0.245 m and a

Question

A bolt of mass 1.90×102 kg moves with SHM that has an amplitude of 0.245 m and a period of 1.485 s . The displacement of the bolt is + 0.245 m when t=0.

Part A

Find the displacement of the bolt at time t = 0.510 s .

Part B

Find the magnitude of the force acting on the bolt at time 0.510 s

Part C

Find the direction of the force acting on the bolt at time t = 0.510 s . (+x -direction or -x -direction)

Part D

What is the minimum time required for the bolt to move from its initial position to the point x = -0.175 m ?

Part E

What is the speed of the bolt at -0.175 m ?

I need this question answered with as soon as possible, within the next 1.5hrs. This is really urgent and would help me understand how to solve such a problem for my exam. Any help/ solution provided would be much aprreciated. Thanks in advance.

Explanation / Answer

here

mass of bolt, m = 0.0190 kg
amplitude, a = 0.245 m

angular velocity, w = 2*pi/T = 2*pi/1.485 = 4.231 rad/s

Part a:
x = A*cos(w*t)

at t = 0.510

x(0.510) = 0.245 * Cos(4.231 * 0.510)
x(0.510) = - 0.136 m

Part b:
acceleration, a = -w^2*A*cos(wt)
acceleration, a = -4.231^2 * 0.245 * cos(4.321 * 0.510 )
acceleration, a = 2.6 m/s^2

Force = mass * acceleration
Force = 0.0190 * 2.6
Force = 0.049 N

Part c:
Since F is + then the force is in the +x direction

Part d:
x = A*cos(w*t)
- 0.175 = 0.235 * Cos(4.231 * t)
4.231 * t = arcCos(-0.175/0.235)
time, t = 0.57 s

Part e:
velocity, v = - w * A * Sin(wt)
velocity, v = - 4.321 * 0.245 * Sin(4.321 * 0.57)
velocity, v = 0.665 m/s

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