A bolt of mass 1.90×10 2 kg moves with SHM that has an amplitude of 0.240 m and
ID: 1510417 • Letter: A
Question
A bolt of mass 1.90×102 kg moves with SHM that has an amplitude of 0.240 m and a period of 1.520 s . The displacement of the bolt is + 0.240 m when t=0.
A) Find the displacement of the bolt at time t = 0.510 s.
B) Find the magnitude of the force acting on the bolt at time 0.51 s.
C) Find the direction of the force acting on the bolt at time t = 0.510 s .
D) What is the minimum time required for the bolt to move from its initial position to the point x = -0.180 m?
E) What is the speed of the bolt at -0.180 m?
Explanation / Answer
x = A cos(2 f t + )
where A is the amplitude, f is the frequency, and is the phase in radians.
The period, T, is the length of time taken by one cycle, and is the reciprocal of the frequency f:
T = 1 / f
In this problem T = 1.520 s = 1 / f
So frequency = 0.658 cycle/s
We are given that A = 0.24 meters
M = mass = 0.019 kg
x = +0.24 meters when t=0
1.) the question is what is x when t = 0.510 s
x = A cos(2 f t + )
x = 0.24 cos[2(3.14)(0.658)(0.51) + ] = 0.24 cos[2.1 + ]
[Note: Now we have two unknowns, but we also know x = +0.24 meters when t=0, which
means 0.24 = 0.24 cos[0 + ], so = 0]
So x = 0.24 cos[2.1 radians] = 0.24 (-0.5) = -0.121 m
Remember: there are radians in 180 degrees.
2.) the question is what is F when t = 0.51 s
F = - k x where F = force in newtons; k = constant
Now T also equals: T = 1 / f = 2 SQRT[M/k]
So 1.520 = 2(3.14) SQRT[0.019 / k]
Therefore k = 0.324 N/m
So F = - k x = - (0.324 N/m)(-0.121 m) = 0.0392 N
3.) from 2.) above the force F is acting in the +x direction, because F is +
4.) the question is what is the minimum time for x = - 0.18 m ?
The first time x = - 0.8 m is:
x = 0.24 cos[2(3.14)(0.658) t] = 0.24 cos[4.13 t]
- 0.18 = 0.24 cos[4.13 t]
cos[4.13 t] = - 0.75
4.13 t = 2.41 radians
t = 0.585 s
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