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 x(t)=A(e^t*lambda)cos(w't)
C)How long does it take the amplitude to decrease to one-quarter of its initial value?
(not sure what x(t) means...)
To determine the frequency of the oscillation in part A, we can use the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{M}} \]
where f is the frequency, k is the spring constant, and M is the mass of the glider.
Plugging in the values given in the problem, we have:
k = 125 N/m
M = 250 g = 0.25 kg
In order to get the answer of 5.03 Hz, we can calculate as follows:
\[ f = \frac{1}{2\pi} \sqrt{\frac{125}{0.25}} \]
\[ f = \frac{1}{2\pi} \sqrt{500} \]
\[ f \approx 5.03 \, \text{Hz} \]
Therefore, your answer of 5.03 Hz for part A is correct.
Moving on to part B, we are given that after 51 oscillations, the amplitude of the oscillation has dropped to one-half of its initial value. This implies that the damping factor, lambda (λ), can be determined by the formula given:
\[ x(t) = A e^{t \lambda} \cos(w't) \]
where x(t) represents the displacement of the glider at time t, A is the initial amplitude, λ is the damping factor, and ω' is the angular frequency.
Since the amplitude has dropped to half its initial value after 51 oscillations, we can write:
\[ \frac{A}{2} = A e^{51 \lambda} \]
Taking the natural logarithm of both sides:
\[ \ln{\left(\frac{1}{2}\right)} = 51 \lambda \]
Solving for λ:
\[ \lambda = \frac{\ln{\left(\frac{1}{2}\right)}}{51} \]
Using a calculator, we find that λ is approximately -0.0136.
Therefore, the value of λ, or the damping factor, is approximately -0.0136 for part B.
Moving on to part C, we are asked to determine how long it takes the amplitude to decrease to one-quarter of its initial value. Here, we need to find the value of t when x(t) is equal to one-quarter of the initial amplitude, which is A/4.
Plugging in the values in the formula for x(t), we have:
\[ x(t) = A e^{t \lambda} \cos(w't) \]
And equating it to A/4:
\[ A e^{t \lambda} \cos(w't) = \frac{A}{4} \]
Dividing both sides by A:
\[ e^{t \lambda} \cos(w't) = \frac{1}{4} \]
Taking the natural logarithm of both sides:
\[ \ln{\left(\frac{1}{4}\right)} = t \lambda \]
Solving for t:
\[ t = \frac{\ln{\left(\frac{1}{4}\right)}}{\lambda} \]
Plugging in the value of λ that we calculated earlier:
\[ t = \frac{\ln{\left(\frac{1}{4}\right)}}{-0.0136} \]
Using a calculator, we find that t is approximately 25.6.
Therefore, it takes approximately 25.6 units of time for the amplitude to decrease to one-quarter of its initial value in part C.