A 60 kg girl on a swing is pulled 1.5 meters back and then released. The girl oscillates at pi/3 radians per second.
How far from equilibrium will the girl be in 4 seconds into the motion?
How fast will she be moving at this location?
2.3 meters in 6 meters per second
To find how far from equilibrium the girl will be after 4 seconds, we need to use the equation for the displacement of an object in simple harmonic motion:
x(t) = A * sin(ωt + φ)
where:
x(t) is the displacement from equilibrium at time t
A is the amplitude (maximum displacement)
ω is the angular frequency (2π times the frequency)
φ is the phase constant
Given that the girl oscillates at pi/3 radians per second, the angular frequency can be calculated as follows:
ω = π/3
The phase constant φ determines the starting position of the girl. Since she is pulled 1.5 meters back, her initial displacement is -1.5 meters:
φ = -1.5
Now we can plug in the values into the equation:
x(t) = A * sin(ωt + φ)
After 4 seconds, the equation becomes:
x(4) = A * sin(π/3 * 4 + -1.5)
To solve for x(4), we need to determine the amplitude A. The amplitude can be calculated using the equation:
A = Maximum displacement - Equilibrium position
In this case, the maximum displacement is 1.5 meters (since the girl is pulled back 1.5 meters), and the equilibrium position is 0 meters. Therefore:
A = 1.5 - 0 = 1.5
Now we can substitute A and solve for x(4):
x(4) = 1.5 * sin(π/3 * 4 + -1.5)
Calculating this value gives us the distance from equilibrium after 4 seconds.
To find the girl's velocity at this location, we can use the equation:
v(t) = ω * A * cos(ωt + φ)
Plugging in the values:
v(4) = π/3 * 1.5 * cos(π/3 * 4 + -1.5)
Calculating this value will give us the speed at the given location.