A particle of mass m = 0.40 kg oscillates on the end of a spring with an amplitude of 0.30 m. At t = 0, it's at it's equilibrium position (x=0) and is moving in the positive x-direction. At 0.20 s later it is at x = 0.25 m.
a. What is the lowest possible value for the spring constant?
b. What is the second lowest possible value for the spring constant?
if 0 at 0 and v+ then
x = .3 sin 2 pi f t
.25 = .3 sin 2 pi f(.2)
.8333 = sin 2 pi f(.2)
so
2 pi f(.2) = sin^-1 .8333 = 56.4 deg = .985 radians
so
2 pi f = 4.92 radians/second
but
w = sqrt(k/m) = 2 pi f
so
k/m = (4.92)^2
k = .4 (4.92)^2
the next one will be at twice the frequency so it makes a second trip in that time
2 pi f = w = 9.84 radians /second
Thanks, I wasn't able to get the second part of the answer (got a spring constant of 24.21 N/m) correct though.
For the second part, since this is a sine function, there's a particular angle that would have the same value as 0.8333. Like sin(0 deg) and sin(180 deg) equals to 0. So I did 180 - 56.4 to get that particular angle. Once I got that, I converted it to radians and repeated the steps you provided after that. Thanks for the help Damon.
To find the spring constant for the given scenario, we can use the equation of motion for simple harmonic motion (SHM):
x(t) = A * cos(ωt + φ)
where:
- x(t) is the displacement of the particle at time t
- A is the amplitude of oscillation
- ω is the angular frequency (related to the spring constant and mass of the particle)
- φ is the phase constant
Let's solve the problem step by step:
a. What is the lowest possible value for the spring constant?
First, let's find the angular frequency ω using the given information.
We know that the particle is at its equilibrium position (x = 0) at t = 0. This means that at t = 0, the particle has not yet started oscillating. Therefore, the phase constant φ = 0.
Using this information, we can rewrite the equation of motion as x(t) = A * cos(ωt).
At t = 0.20 s, the particle is at x = 0.25 m.
Substituting these values into the equation, we have:
0.25 = 0.30 * cos(ω * 0.20)
To find the value of ω, we need to solve this equation for ω. Here are the steps:
1. Divide both sides of the equation by 0.30:
0.25 / 0.30 = cos(ω * 0.20)
2. Take the inverse cosine (cos^(-1)) of both sides:
cos^(-1)(0.25 / 0.30) = ω * 0.20
3. Solve for ω:
ω = cos^(-1)(0.25 / 0.30) / 0.20
Using a calculator, we find:
ω ≈ 0.9381 rad/s
Now, we can relate ω to the spring constant k and the mass m using the equation:
ω = √(k / m)
Squaring both sides of the equation and isolating k, we get:
k = ω^2 * m
Substituting the values of ω and m, we can find the lowest possible value for the spring constant:
k = (0.9381 rad/s)^2 * 0.40 kg
Calculating this, we find:
k ≈ 0.351 N/m
Therefore, the lowest possible value for the spring constant is approximately 0.351 N/m.
b. What is the second lowest possible value for the spring constant?
To find the second lowest possible value for the spring constant, we need to consider the next possible phase constant φ.
In the previous calculation, we assumed that φ = 0, which means the particle started at its equilibrium position. Now, let's consider the case where the particle starts at the maximum displacement, A, instead.
When the particle starts at its maximum displacement, it reaches x = 0.25 m at t = 0 (instead of t = 0.20 s in the previous case).
Using the equation x(t) = A * cos(ωt), and substituting x = 0.25 m and t = 0, we have:
0.25 = A * cos(ω * 0)
Since cos(0) = 1, we get:
0.25 = A * 1
This means A = 0.25 m.
Now, we can find the angular frequency ω using the equation ω = √(k / m) as before. We can substitute A = 0.25 m and m = 0.40 kg into the equation to solve for ω.
Next, using the equation k = ω^2 * m, we can find the second lowest possible value for the spring constant k.
Following these steps, we can calculate the second lowest possible value for the spring constant.