A 2.24 kg mass on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.45 N/m. The mass is displaced 3.06 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. What is the force (including direction, choose the right to be positive) acting on the mass 4.20 s after it is released
To determine the force acting on the mass 4.20 seconds after it is released, we first need to find the position and velocity of the mass at that time.
Since the mass is undergoing simple harmonic motion (SHM), we can use the equations of SHM to find the position and velocity at any given time.
The equation for the displacement (x) of an object undergoing SHM is given by:
x = A * cos(ωt + φ)
Where:
- A is the amplitude of the motion (maximum displacement from equilibrium position)
- ω is the angular frequency of the motion
- t is the time
- φ is the phase constant (initial phase angle)
In this case, the mass is initially displaced 3.06 m to the right from the equilibrium position, so the amplitude (A) is 3.06 m.
The angular frequency (ω) can be found using the equation:
ω = √(k / m)
Where:
- k is the force constant of the spring (4.45 N/m)
- m is the mass of the object (2.24 kg)
Plugging in the values, we have:
ω = √(4.45 N/m / 2.24 kg) = √1.984375 N/kg ≈ 1.4104 rad/s
Now, let's find the phase constant (φ). At t = 0, the mass is released from its initial displacement to initiate the motion. At that moment, the object is at its maximum displacement to the right. Therefore, the phase constant is 0, since the cosine of 0 is 1.
Now we can use the equation to find the position at t = 4.20 seconds:
x = 3.06 m * cos(1.4104 rad/s * 4.20 s + 0)
x ≈ 3.06 m * cos(5.9203 rad)
x ≈ 3.06 m * (-0.9228)
x ≈ -2.8264 m
So, the mass is at a displacement of approximately -2.8264 m from the equilibrium position after 4.20 seconds.
Next, we need to find the velocity of the mass at that time. The velocity (v) of an object undergoing SHM is given by:
v = -A * ω * sin(ωt + φ)
Plugging in the values, we have:
v = -3.06 m * 1.4104 rad/s * sin(1.4104 rad/s * 4.20 s + 0)
v ≈ -3.06 m * 1.4104 rad/s * sin(5.9203 rad)
v ≈ -3.06 m * 1.4104 rad/s * 0.5651
v ≈ -2.9658 m/s
The negative sign indicates that the velocity is directed to the left.
Now, we can calculate the force acting on the mass using Newton's second law of motion, which states that the force (F) acting on an object is equal to its mass (m) multiplied by its acceleration (a):
F = m * a
Since the mass is undergoing SHM, the acceleration (a) at any given time is given by:
a = -ω^2 * x
Plugging in the values, we have:
a = -1.4104 rad/s^2 * (-2.8264 m)
a ≈ 3.9809 m/s^2
Finally, we can calculate the force at t = 4.20 seconds:
F = 2.24 kg * 3.9809 m/s^2
F ≈ 8.9195 N
Therefore, the force acting on the mass 4.20 seconds after it is released is approximately 8.9195 N, directed to the left.