For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 n/m. the damping force is given by -b(dx/dt), where b = 210 g/s. the block is pulled down 13.0 cm and released. (a) calculate the time required for the amplitude of the resulting oscillations to fall to 1/8 of its initial value. (b) how many oscillations are made by the block in this time?

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

where A(t)=A₀

 A₀ is the amplitude at t=0 and

is the angular frequency of damped SHM, which is given by,

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 =A₀/8

⇒

applying logarithm on both sides

⇒

⇒

substituting the values

b)

, where is time period of damped SHM

⇒

let be number of oscillations made

then,

⇒

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas