Question

A 250 g air-track glider is attached to a spring with springconstant 4.0 N/m. Th damping constant due to air resistance is0.015 kg/s. The glider is pulled out 20cm from equilibrium and released. How many oscillations will it make during the time which the amplitude decays to e-1 of tis initial value?

Answer 1

Answer:

33.33 seconds

Explanation:

$N=\dfrac{1}{e}N_0$

$N_0$ = Initial length pulled = 20 cm

b = Damping constant = 0.015 kg/s

k = Spring constant = 4 N/m

m = Mass of glider = 250 g

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.25}{4}}\\\Rightarrow T=1.57079632679\ s$

Using exponential decay formula

$N=N_0e^{\dfrac{-bt}{m}}$

Final amplitude = Initial times decay

$\dfrac{1}{e}0.2=0.2e^{\dfrac{-0.015t}{2\times 0.25}}\\\Rightarrow 0.2=0.2e^{\frac{-0.015t}{2\cdot \:0.25}+1}\\\Rightarrow e^{\frac{-0.015t}{2\cdot \:0.25}+1}=1\\\Rightarrow \ln \left(e^{\frac{-0.015t}{2\cdot \:0.25}+1}\right)=\ln \left(1\right)\\\Rightarrow \left(\frac{-0.015t}{2\cdot \:0.25}+1\right)\ln \left(e\right)=\ln \left(1\right)\\\Rightarrow \frac{-0.015t}{2\cdot \:0.25}+1=\ln \left(1\right)\\\Rightarrow -\frac{0.015t}{0.5}=-1\\\Rightarrow -0.000225t=-0.0075\\\Rightarrow t=33.33\ s$

The time taken is 33.33 seconds