A mass attached to a spring oscillates with a period of 3.15 s. If the mass starts from rest at x= 4.10×10−2 m and time t= 0, where is it at time t= 6.37s?
Is the mass moving in the positive or negative x direction at t= 6.37s?
To determine the position of the mass at time t = 6.37 s, we need to use the equation of motion for simple harmonic motion:
x(t) = A * cos(ωt + φ)
Where:
x(t) is the position of the mass at time t
A is the amplitude of the oscillation
ω is the angular frequency (2π / T), where T is the period of the oscillation
φ is the phase constant
First, we need to find the angular frequency ω using the formula:
ω = 2π / T
Given the period T = 3.15 s:
ω = 2π / 3.15 = 1.995 rad/s
Next, we can substitute the given values into the equation of motion to find the phase constant φ:
At t = 0, x = 4.10 × 10^(-2) m
x(0) = A * cos(φ) = 4.10 × 10^(-2) m
We can rearrange this equation to solve for the amplitude A:
A = x(0) / cos(φ) = 4.10 × 10^(-2) m / cos(φ)
Now, we have the amplitude A and the angular frequency ω, so we can find the position of the mass at time t = 6.37 s:
x(6.37) = A * cos(ω * t) = A * cos(1.995 * 6.37)
Substituting the values:
x(6.37) = (4.10 × 10^(-2) m / cos(φ)) * cos(1.995 * 6.37)
Simplifying this equation will give us the position of the mass at time t = 6.37 s.
To determine if the mass is moving in the positive or negative x direction at t = 6.37 s, we need to evaluate the sign of the position at that time. If x(6.37) is positive, the mass is moving in the positive x direction. If x(6.37) is negative, the mass is moving in the negative x direction.