A glider on an air track is connected by springs to either end of the track. Both springs have the same spring constant,k , and the glider has mass M.
A)Determine the frequency of the oscillation, assuming no damping, if k= 125N/m and M= 250g. (I got the answer to be 5.03 hz)
B)It is observed that after 51 oscillations, the amplitude of the oscillation has dropped to one-half of its initial value. Estimate the value of (lambda), using A(e^t*lambda)cos(w't)
C)How long does it take the amplitude to decrease to one-quarter of its initial value?
Fortnite
equivalent spring constant = 2k = 250 N/m
w = sqrt(k/m) = srt(250/.25) = sqrt(1000)=10 sqrt 10
2 pi f = w = 10 sqrt 10
f = (10/2pi)sqrt 10 = 5.03 check
e^Lt = .5
T = 1/f = .1987 seconds
51 T = 10.13 s
so
e^L(10.13) = .5
10.13 L = ln .5 = -.693
L = -.0684
Now you have L
e^-.0684 t = .25
etc
but how do you find A (i'm assuming amplitude)?
It did not ask for A. You can not find A. All you can find is the ratio of A to the original A.
A/Ao = e^Lt
the equation to use is more specifically x(t)=A(e^t*L)cos(w't)
does that change it? because i'm still not getting the right answer
The answer is right but it should be positive since the formula as I know it is e^(-L*t)
A) To determine the frequency of the oscillation, we can use the equation:
f = (1/2π) * √(k/M)
Given that k = 125 N/m and M = 250 g, we need to convert the mass to kilograms:
M = 250 g = 0.25 kg
Thus, the frequency is:
f = (1/2π) * √(125/0.25) = (1/2π) * √(500) ≈ 5.03 Hz
Your answer is correct.
B) To estimate the value of λ when the amplitude drops to one-half of its initial value after 51 oscillations, we can use the equation you provided:
A(t) = A(e^(-t * λ)) * cos(w' * t)
Given that the amplitude decreases to one-half, we can substitute A(t) with A/2 and solve for λ. Let's assume the initial amplitude is A₀.
A/2 = A₀ * (e^(-51 * λ))
Taking the natural logarithm (ln) of both sides, we get:
ln(A/2) = -51 * λ * ln(e)
Since ln(e) is equal to 1, we have:
ln(A/2) = -51 * λ
Simplifying further:
λ = -ln(A/2) / 51
Plug in the correct values and you will get the estimation for λ.
C) To determine how long it takes for the amplitude to decrease to one-quarter of its initial value, we can again use the equation:
A(t) = A(e^(-t * λ)) * cos(w' * t)
Now we need to solve for t when A(t) is equal to one-quarter of the initial amplitude. Let's assume the initial amplitude is A₀.
A(t) = A₀ * (e^(-t * λ)) = (1/4) * A₀
Divide both sides by A₀:
e^(-t * λ) = 1/4
Taking the natural logarithm of both sides:
-ln(1/4) = -t * λ
Simplifying further:
t * λ = ln(4)
t = ln(4) / λ
Plug in the value you estimated for λ earlier, and you will get the time it takes for the amplitude to decrease to one-quarter of its initial value.