Question

For a damped harmonic oscillation, the equation of motion is qquad m(d^(2)x)/(dt^(2))+gamma(dx)/(dt)+kx=0 with m=0.20kg, gamma=0.04kg^(-1) and k=65Nm^(-1) .Calculate (i) the period of motion, (ii) number of oscillations in which its amplitude will become haf of its initial value ,and (iiii) the number of oscillations in which its mechanical energy will drop to half of its initial value.​

Answer 1

Answer:

i) time period of oscillation is 0.348 s

ii) It will have 20 number of oscillations during the time its amplitude becomes half

iii) It will have 10 number of oscillations during the time its amplitude becomes half

Explanation:

Part a)

Time period of the oscillation of the pendulum is given as

$T = 2\pi\sqrt{\frac{m}{k}}$

here we have

m = 0.20 kg

k = 65 N/m

so we will have

$T = 2\pi\sqrt{\frac{0.20}{65}}$

$T = 0.348 s$

Part b)

As we know that amplitude of SHM is given as

$A = A_oe^{-bt/2m}$

now as the amplitude becomes half then we have

$0.5 = e^{-bt/2m}$

$t = \frac{2m}{b} ln2$

$t = \frac{2(0.20)}{0.04} ln2$

$t = 6.93 s$

So total number of oscillations in this time interval is given as

$N = \frac{t}{T}$

$N = \frac{6.93}{0.348} = 20$

Part c)

Mechanical energy of damped oscillation is given as

$E = E_o e^{-bt/m}$

now energy becomes half of initial energy then we have

$0.5 = e^{-bt/m}$

$t = \frac{m}{b} ln 2$

$t = \frac{0.20}{0.04} ln 2$

$t = 3.46 s$

Now number of oscillations are given as

$N = \frac{t}{T}$

$N = \frac{3.46}{0.348} = 10$

#Learn

Topic : Damped oscillations

https://brainly.in/question/12222621