The amplitude of a damped oscillator decreases to 0.9 times its original value in 5 s. In another 10 s it will decreases to…

Q: The amplitude of a damped oscillator decreases to 0.9 times its original value in 5 s. In another 10 s it will decreases to α times its original magnitude, where α is

Sol: $\large A = A_0 e^{-\frac{b t}{2m}}$

after 5 sec⁡ ,

$\large 0.9 A_0 = A_0 e^{-\frac{5 b}{2m}}$

$\large 0.9 = e^{-\frac{5 b}{2m}}$

After 10 more sec (i.e., t = 15 sec) its amplitude becomes α A0.

$\large \alpha A_0 = A_0 e^{-\frac{15 b}{2m}}$

$\large \alpha = e^{-\frac{15 b}{2m}}$

$\large \alpha = (e^{-\frac{5 b}{2m}})^3 = (0.9)^3$

= 0.729

A mass of 2 kg oscillates on a spring with force constant 50 N/m. By what factor does the frequency of oscillation decrease..

Q: A mass of 2 kg oscillates on a spring with force constant 50 N/m. By what factor does the frequency of oscillation decrease when a damping force with constant b = 12 is introduced ?

sol: $\large \omega_0 = \sqrt{\frac{50}{2}} = 5 $

$\large \omega = \sqrt{\omega_0^2 – (\frac{b}{2m})^2}$

$\large \omega = \sqrt{5^2 – 3^2} = 4 $

So , frequency reduces by 1 Hz or 20 %.

The trolley car having simple pendulum decelerated by friction. In consequence, the pendulum has time period T…

Q: The trolley car having simple pendulum decelerated by friction. In consequence, the pendulum has time period T. If T0 is time period of the simple pendulum in the absence of any acceleration of the trolley car, the value of T/T0 is …

Sol: $\large T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + a^2}}}$

a = μ g

$\large T = 2\pi \sqrt{\frac{l}{g \sqrt{1 + \mu^2}}}$

$\large T = T_0 \sqrt{\frac{1}{\sqrt{1 + \mu^2}}}$

$\large \frac{T}{T_0} = \sqrt{\frac{1}{\sqrt{1 + \mu^2}}}$

A Simple pendulum has time period ‘T1’. The point of suspension is now moved upwards according to the relation y = kt² …

Q: A Simple pendulum has time period ‘T1’. The point of suspension is now moved upwards according to the relation y = kt², (k = 1 m/sec²) where y is the vertical displacement. The time period now becomes ‘ T2 ’, then find the ratio of  $\frac{T_1^2}{T_2^2}$

Sol: y = k t2

$\large \frac{dy}{dt} = 2 k t $

$\large \frac{d^2y}{dt^2} = 2 k = 2 $

a = 2 m/s2 (acceleration)

$\large T_1 = 2\pi \sqrt{\frac{l}{g}}$

$\large T_2 = 2\pi \sqrt{\frac{l}{g+a}}$

$\large \frac{T_1^2}{T_2^2} = \frac{g+a}{g}$

$\large \frac{T_1^2}{T_2^2} = \frac{10+2}{10} = \frac{6}{5}$

A block of mass ‘m’ is attached to the light spring of force constant K and released when it is in its natural length…

Q: A block of mass ‘m’ is attached to the light spring of force constant K and released when it is in its natural length. Find amplitude of subsequent oscillations.

Sol: From conservation of energy $\large m g x = \frac{1}{2} K x^2 $

The maximum displacement of the spring in subsequent motion will be 2mg/K

From F = Kx, mg = Kx The equilibrium position of the system will occur at the extension of mg/K

amplitude $\large = \frac{2mg}{K} – \frac{mg}{K} = \frac{mg}{K} $

A particle of mass 10g executes a linear SHM of amplitude 5cm with a period of 2 s. Find the PE and KE…

Q: A particle of mass 10g executes a linear SHM of amplitude 5cm with a period of 2 s. Find the PE and KE, 1/6 s after crossing the mean position.

Sol: Given m = 10 g = 10-2 kg ,T = 2 s,

ω = 2π/T = 2π/2 = π rad/s

A = 5 cm = 5 × 10-2 m

$\large K.E = \frac{1}{2}m \omega^2 A^2 cos^2 \omega t $

At t=1/6 s ,

$\large K.E = \frac{1}{2}\times 10^{-2} \pi^2 (5 \times 10^{-2})^2 cos^2 \pi/6 $

= 9.25 × 10-5 J

$\large P.E = \frac{1}{2}m \omega^2 A^2 sin^2 \omega t $

$\large P.E = \frac{1}{2}\times 10^{-2} \pi^2 (5 \times 10^{-2})^2 sin^2 \pi/6 $

= 3.085 × 10-5 J