A 0.55 kg object connected to a light spring with a force constant of 20.0 N/m oscillates on a frictionless horizontal surface. If the spring is compressed 4.0 cm and released from rest.
(a) Determine the maximum speed of the object.
(b) Determine the speed of the object when the spring is compressed 1.5 cm.
2 cm/s
(c) Determine the speed of the object when the spring is stretched 1.5 cm.
3 cm/s
(d) For what value of x does the speed equal one-half the maximum speed?
To solve these problems, we need to use the equations of motion for simple harmonic motion. Let's go step by step to find the answers.
(a) To determine the maximum speed of the object, we can use the equation v_max = Aω, where A is the amplitude of the oscillation and ω is the angular frequency. The angular frequency can be calculated using the formula ω = √(k/m), where k is the force constant of the spring and m is the mass of the object.
Given that the spring is compressed by 4.0 cm, which is equivalent to 0.04 m, and the force constant k is 20.0 N/m, and the mass, m, is 0.55 kg, we can find the amplitude:
A = 0.04 m
k = 20.0 N/m
m = 0.55 kg
Substituting these values into the equation for the angular frequency, we get:
ω = √(k/m) = √(20.0 N/m / 0.55 kg) ≈ 6.710 s^(-1)
Now, we can substitute the values of A and ω into the equation for the maximum speed:
v_max = Aω = (0.04 m) * (6.710 s^(-1)) ≈ 0.2684 m/s
Therefore, the maximum speed of the object is approximately 0.2684 m/s.
(b) To determine the speed of the object when the spring is compressed 1.5 cm, or 0.015 m, we can use the same formula:
v = Aω
Substituting the value of A and ω we calculated in part (a), we get:
v = (0.015 m) * (6.710 s^(-1)) ≈ 0.1007 m/s
Therefore, the speed of the object when the spring is compressed 1.5 cm is approximately 0.1007 m/s.
(c) To determine the speed of the object when the spring is stretched 1.5 cm, we can again use the same formula:
v = Aω
Since stretching the spring is essentially the same as compressing it, the speed will be the same as in part (b). Therefore, the speed of the object when the spring is stretched 1.5 cm is also approximately 0.1007 m/s.
(d) To find the value of x when the speed is equal to one-half the maximum speed, we can use the formula for speed:
v = Aω
We can rearrange this equation to solve for A:
A = v/ω
We can then substitute the given values:
v = 0.5 * (0.2684 m/s) = 0.1342 m/s
ω = √(k/m) = √(20.0 N/m / 0.55 kg) ≈ 6.710 s^(-1)
Substituting these values into the equation for A, we get:
A = (0.1342 m/s) / (6.710 s^(-1)) ≈ 0.01999 m
Therefore, when the speed is equal to one-half the maximum speed, the value of x is approximately 0.01999 m.