Simple Harmonic Motion

The formula for the period of a simple pendulum is given by:

$ T = 2\pi\sqrt{\frac{l}{g}} $

where:

• T is the period of the pendulum,
• l is the length of the pendulum, and
• g is the acceleration due to gravity.

We need to find g. Rearranging the formula for g, we get:

$ g = \frac{4\pi^2l}{T^2} $

Given:

• l = 2 m,
• T = 2 s,

Substituting these values into the equation, we get:

$ g = \frac{4\pi^2 \times 2}{(2)^2} = 2\pi^2 \, ms^{-2} $

Potential of the given oscillator is

$V = {1 \over 2}k{(x - k)^2}$

Given: M = 10; m = 5, u = 1; k = 1

Initial momentum of the particle of mass m

= mu = m $\times$ 5 = 5m

Momentum of (oscillator + particle) after collision = (M + m)

Velocity of oscillator after collision = v

So, momentum of system = (M + m)v

From conservation of linear momentum, we have

(M + m) = mu = 5 $\times$ 1 = 5

For second collision, oscillator and particle have momentum in opposite direction.

Net or total momentum is zero.

Likewise after 4^th, 6^th, 8^th, 10^th, 12^th collision the momentum is zero. After 12^th collision, Mass of oscillator and 12 particles will be (10 + 12 $\times$ 5) = 70

Now, from conservation of linear momentum, for 13^th collision, we have

$70 \times 0 + 5 \times 1 = (70 + 5)v' \Rightarrow v' = {5 \over {75}} \Rightarrow {1 \over {15}}$

Total mass after 13^th collision = (10 + 13 $\times$ 5) = 75

Kinetic energy of system $ = {1 \over 2}mv{'^2}$

$ \Rightarrow KE = {1 \over 2} \times 75 \times {1 \over {15}} \times {1 \over {15}}$

$ \Rightarrow {1 \over 2}k{A^2} = {1 \over 2} \times {{75} \over {225}} = {1 \over 6}$

$ \Rightarrow {1 \over 2} \times 1 \times {A^2} = {1 \over 6} \Rightarrow {A^2} = {1 \over 3}$

$ \Rightarrow A = {1 \over {\sqrt 3 }}$

The given simple harmonic motions to form Lissajous figures are $x(t) = A\sin (at + \delta )$ and $y(t) = B\sin (bt)$.

For parabola, conditions should be

A = B or A $\ne$ B, a = 2b, $\delta$ = $\pi$/2

For line, conditions should be

A = B, a = b, $\delta$ = $\pi$

For circle, condition should be

A = B, a = b; $\delta$ = $\pi$/2

For ellipse, condition should be

A $\ne$ B, a = b; $\delta$ = $\pi$/2

Therefore, we obtain an ellipse.

To determine the time taken for the mechanical energy of the damped oscillator to drop to half its initial value, we'll use the principles of damped harmonic motion.

Given:

Mass of the block ($ m $): 0.1 kg

Spring constant ($ k $): 640 N/m

Damping constant ($ b $): $ 10^{-2} $ kg/s

Understanding Damped Harmonic Motion:

In damped harmonic motion, the amplitude of oscillation decreases exponentially over time due to the damping force. The mechanical energy ($ E $) of the oscillator is proportional to the square of its amplitude ($ A $):

$ E(t) \propto A(t)^2 $

The amplitude as a function of time is given by:

$ A(t) = A_0 \, e^{- \frac{b}{2m} t} $

Where:

$ A_0 $ is the initial amplitude.

$ b $ is the damping constant.

$ m $ is the mass.

$ t $ is the time.

Therefore, the mechanical energy as a function of time is:

$ E(t) = E_0 \, e^{- \frac{b}{m} t} $

Where $ E_0 $ is the initial mechanical energy.

Calculating the Time When Energy Drops to Half:

We need to find the time $ t $ when $ E(t) = \frac{1}{2} E_0 $:

$ \frac{1}{2} E_0 = E_0 \, e^{- \frac{b}{m} t} $

Simplify:

$ \frac{1}{2} = e^{- \frac{b}{m} t} $

Take the natural logarithm of both sides:

$ \ln\left(\frac{1}{2}\right) = - \frac{b}{m} t $

Simplify $ \ln\left(\frac{1}{2}\right) = -\ln(2) $:

$ - \ln(2) = - \frac{b}{m} t $

Cancel negatives:

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

Solve for $ t $:

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

Plugging in the Given Values:

$ t = \frac{0.1\, \text{kg}}{10^{-2}\, \text{kg/s}} \ln(2) $

Calculate $ \ln(2) $:

$ \ln(2) \approx 0.6931 $

Now compute $ t $:

$ t = \frac{0.1}{0.01} \times 0.6931 = 10 \times 0.6931 = 6.931\, \text{s} $

Conclusion:

The time taken for the mechanical energy to drop to half its initial value is approximately 6.93 seconds, which is closest to 7 seconds among the given options.