A spring with a mass attached has a natural length at rest of 20 cm. The mass is pulled forward 8 cm and released. The spring oscillates 1 time in 4
seconds. Absent friction, find the times the mass is 3 cm behind the resting
location in the first 8 seconds.
To find the times when the mass is 3 cm behind the resting location in the first 8 seconds, we need to understand the properties of the oscillating mass-spring system.
First, we can determine the amplitude of the oscillation. The amplitude is the maximum displacement from the equilibrium position. In this case, the mass is pulled forward 8 cm, so the amplitude is 8 cm.
Next, we can find the period of the oscillation. The period is the time taken to complete one full oscillation. We are given that the spring oscillates once in 4 seconds, so the period is 4 seconds.
The formula for the period of an oscillating mass-spring system is:
T = 2π√(m/k),
where T is the period, m is the mass, and k is the spring constant.
We are not given the mass or the spring constant, but we can find them using the information provided. Since the mass is pulled forward and released, it is experiencing simple harmonic motion, and we can use the relationship:
T = 2π√(m/k) = 4 seconds.
Squaring both sides of the equation, we get:
16π²(m/k) = 16,
which simplifies to:
π²(m/k) = 1.
Since we are not given the mass or the spring constant, we cannot solve for them explicitly. However, we can use the information given in the question to find the ratio m/k.
The natural length of the spring at rest is 20 cm, and the mass is pulled forward 8 cm. When the mass is released, it will oscillate around the equilibrium position, which is now 8 cm ahead of the natural length.
Therefore, the maximum displacement from the equilibrium position is 8 cm, which corresponds to the amplitude of the oscillation. Using the formula for the amplitude of a mass-spring system:
A = (m/k) √(k/m),
we can substitute the known values:
8 cm = (m/k) √(k/m).
Squaring both sides of the equation and rearranging terms, we get:
(m/k)² = 64.
Combining this equation with the previous equation π²(m/k) = 1, we can solve for the ratio m/k:
π²(m/k)² = π² × 64,
(m/k)² = 64/π²,
m/k = √(64/π²),
m/k = 8/π.
Now that we have the ratio m/k, we can proceed to find the times when the mass is 3 cm behind the resting location in the first 8 seconds.
The equation that describes the displacement of an oscillating mass-spring system as a function of time is:
x(t) = A cos(2πt/T),
where x(t) is the displacement at time t, A is the amplitude, and T is the period.
In this case, the amplitude A is 8 cm, and the period T is 4 seconds.
Substituting these values into the equation, we get:
x(t) = 8 cos(2πt/4),
x(t) = 8 cos(πt/2).
Now, we want to find the times t when the displacement x(t) is 3 cm behind the resting location, which is 20 cm. Therefore, we have:
x(t) = 20 - 3,
8 cos(πt/2) = 17.
To find the values of t that satisfy this equation, we can solve for t by taking the inverse cosine (also known as the arccos) of both sides:
πt/2 = arccos(17/8),
t = (2/arccos(17/8)) × π.
Once we have the value of t, we can check whether it falls within the first 8 seconds as given in the question. If it does, then that is one of the times when the mass is 3 cm behind the resting location.
Repeating this process, we can find other values of t that satisfy the equation for x(t) = 20 - 3, within the first 8 seconds.