A glider on an air track is connected by springs to either end of the

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.