Oscillations of air-track glider

[Kalie]

A 160 g air-track glider is attached to a spring with spring constant 4.40 N/m. The damping constant due to air resistance is 2.40×10−2 kg/s. The glider is pulled out 23.0 cm from equilibrium and released.

How many oscillations will it make during the time in which the amplitude decays to e^-1 of its initial value?

I have no clue how to approach this problems except that
x(t)=Ae^-bt/2m=Ae^-t/2tau
but how can I fin that amount of oscillations?
I'm confused and the dead lines in an hour
please help!

 

[OlderDan]

A 160 g air-track glider is attached to a spring with spring constant 4.40 N/m. The damping constant due to air resistance is 2.40×10−2 kg/s. The glider is pulled out 23.0 cm from equilibrium and released.

How many oscillations will it make during the time in which the amplitude decays to e^-1 of its initial value?

I have no clue how to approach this problems except that
x(t)=Ae^-bt/2m=Ae^-t/2tau
but how can I fin that amount of oscillations?
I'm confused and the dead lines in an hour
please help!

The damped oscillator equation for x(t) has a decaying exponential part and an oscillatory part (sine or cosine). You need both parts to do this problem.

http://hyperphysics.phy-astr.gsu.edu/Hbase/oscda.html

 

[kyle8921]

I would approach this problem by first understanding the physical principles involved. The air-track glider is undergoing simple harmonic motion, meaning that its position over time can be described by a sinusoidal function. The spring constant and damping constant affect the amplitude and frequency of the oscillations.

To solve for the number of oscillations, we need to find the time it takes for the amplitude to decay to e^-1 of its initial value. This is known as the "damping time" and can be calculated using the damping constant and the mass of the glider.

The equation for the damping time is given by t_damp = 2m/b, where m is the mass of the glider and b is the damping constant. Plugging in the values given in the problem, we get t_damp = 2(0.160 kg)/(2.40x10^-2 kg/s) = 13.33 seconds.

Next, we can use the equation for the position of a damped oscillator, x(t) = A0e^(-bt/2m)cos(ωt + φ), where A0 is the initial amplitude, ω is the angular frequency, and φ is the phase angle. We know that the initial amplitude is 23 cm and the damping time is 13.33 seconds. We can also calculate the angular frequency using the spring constant and mass of the glider: ω = √(k/m) = √(4.40 N/m / 0.160 kg) = 6.24 rad/s.

Now, we can plug in these values and solve for the number of oscillations using the equation for e^-1: e^-1 = 1/e = 0.368. This means that after 13.33 seconds, the amplitude of the oscillations will have decayed to 0.368 of its initial value.

We can solve for the number of oscillations by setting the amplitude equal to 0.368A0 and solving for the time t. This gives us:

0.368A0 = A0e^(-bt/2m)cos(ωt + φ)

0.368 = e^(-bt/2m)cos(ωt + φ)

Taking the natural logarithm of both sides and rearranging, we get:

ln(0.368) = -bt/2m + ln(cos(ωt + φ))