A 265g mass attached to a horizontal spring oscillates at a frequency of 1.00Hz . At t =0s, the mass is at x= 4.20cm and has vx = − 1*.0cm/s. Determine the position at t= 3.60s.
So I already found that...
The period is 1.0000 s
The angular frequency is 6.2832 rad/s
The amplitude is 5.1753×10^−2 m
The phase constant is 0.6240 rad
The maximum speed is 0.325 m/s
The maximum acceleration 2.04 m/s^2
The total energy is 1.40×10^−2 J
The position at t= 3.60 s is….????
nvm I figured it out...
the answer was -1.6*10^-2 m
Youve got the wrong amplitude
To determine the position at t = 3.60s, we can use the equation for simple harmonic motion:
x(t) = A cos(ωt + φ),
where:
x(t) is the position at time t,
A is the amplitude,
ω is the angular frequency (ω = 2πf, where f is the frequency),
φ is the phase constant.
We already have the values for A, ω, and φ:
Amplitude (A) = 5.1753 × 10^(-2) m
Angular frequency (ω) = 6.2832 rad/s
Phase constant (φ) = 0.6240 rad
Now, substitute these values into the equation to find the position at t = 3.60s:
x(t) = A cos(ωt + φ)
= (5.1753 × 10^(-2) m) * cos((6.2832 rad/s)(3.60 s) + 0.6240 rad)
To calculate this, first multiply ω and t:
(6.2832 rad/s)(3.60 s) = 22.6603 rad
Then, add the result to the phase constant:
22.6603 rad + 0.6240 rad = 23.2843 rad
Finally, substitute these values back into the equation:
x(t) = (5.1753 × 10^(-2) m) * cos(23.2843 rad)
Using a calculator to find the cosine of 23.2843 rad, we get:
x(t) ≈ (5.1753 × 10^(-2) m) * 0.9952
≈ 5.1425 × 10^(-2) m
Therefore, the position at t = 3.60s is approximately 5.1425 × 10^(-2) m.