Respuesta :
Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀[tex]e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)[/tex]
where A(t)=A₀[tex]e^{\frac{-bt}{2m}}[/tex]
A₀ is the amplitude at t=0 and
[tex]w'[/tex] is the angular frequency of damped SHM, which is given by,
[tex]w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }[/tex]
Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀[tex]e^{\frac{-bt}{2m}}[/tex] =A₀/8
⇒[tex]e^{\frac{bt}{2m}}=8[/tex]
applying logarithm on both sides
⇒[tex]\frac{bt}{2m}=ln(8)[/tex]
⇒[tex]t=\frac{2m*ln(8)}{b}[/tex]
substituting the values
[tex]t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)[/tex]
b) [tex]w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }[/tex]
[tex]w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}[/tex]
[tex]T'=\frac{2\pi}{w'}[/tex], where [tex]T'[/tex] is time period of damped SHM
⇒[tex]T'=\frac{2\pi}{2.86}=2.2s[/tex]
let [tex]n[/tex] be number of oscillations made
then, [tex]nT'=t[/tex]
⇒[tex]n=\frac{24}{2.2}=11(approx)[/tex]
