Simple Harmonic Motion

1st Particle

P = 0 at x = a $\Rightarrow$ 'a' is the amplitude of oscillation 'A₁'.

At x = 0, P = b (at mean position)

$ \Rightarrow m{v_{\max }} = b \Rightarrow {v_{\max }} = {b \over m}$

${E_1} = {1 \over 2}mv_{\max }^2 = {m \over 2}{\left[ {{b \over m}} \right]^2} = {{{b^2}} \over {2m}}$

${A_1}{\omega _1} = {v_{\max }} = {b \over m}$

$ \Rightarrow {\omega _1} = {b \over {ma}} = {1 \over {m{n^2}}}(A = a,\,{a \over b} = {n^2})$

2nd Particle

P = 0 at x = R $\Rightarrow$ A₂ = R

At x = 0, P = R $\Rightarrow$ ${v_{\max }} = {R \over m}$

${E_2} = {1 \over 2}mv_{\max }^2 = {m \over 2}{\left[ {{R \over m}} \right]^2} = {{{R^2}} \over {2m}}$

${A_2}{\omega _2} = {R \over m} \Rightarrow {\omega _2} = {R \over {mR}} = {1 \over m}$

(b) ${{{\omega _2}} \over {{\omega _1}}} = {{1/m} \over {1/m{n^2}}} = {n^2}$

(c) ${\omega _1}{\omega _2} = {1 \over {m{n^2}}} \times {1 \over m} = {1 \over {{m^2}{n^2}}}$

(d) ${{{E_1}} \over {{\omega _1}}} = {{{b^2}/2m} \over {1/m{n^2}}} = {{{b^2}{n^2}} \over 2} = {{{a^2}} \over {2{n^2}}} = {{{R^2}} \over 2}$

${{{E_2}} \over {{\omega _2}}} = {{{R^2}/2m} \over {1/m}} = {{{R^2}} \over 2}$

$ \Rightarrow {{{E_1}} \over {{\omega _1}}} = {{{E_2}} \over {{\omega _2}}}$

The sharpness of resonance decreases with increasing column length.

The first resonance happens when length of air column $\approx$ $\lambda/2$. The prongs of the tuning fork are always kept in a vertical plane and amplitude of vibration is typically in the mm range.

From the figure, we have

${\lambda \over 4} = {l_1} + e$

where e is the end-correction. Therefore,

${l_1} + \left( {{\lambda \over 4} - e} \right) < {\lambda \over 4}$

$ \begin{aligned} & x=A \sin ^2 \omega t+B \cos ^2 \omega t+C \sin \omega t \cos \omega t \\ & x=\frac{A}{2}(1-\cos 2 \omega t)+\frac{B}{2}(1+\cos 2 \omega t)+\frac{C}{2} \sin 2 \omega t \\ & x=\frac{A+B}{2}+\left(\frac{B}{2}-\frac{A}{2}\right)+\cos 2 \omega t+\frac{C}{2} \sin 2 \omega t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} & x=\frac{A+B}{2} \sqrt{\left(\frac{B-A}{2}\right)^2+\left(\frac{C}{2}\right)^2} \sin (2 \omega t+\alpha) \\ & \qquad \tan \alpha=\frac{B-A}{C} \\ & x-\frac{A+B}{2}=\sqrt{\left(\frac{B-A}{2}\right)^2+\left(\frac{C}{2}\right)^2} \sin (2 \omega t+\alpha) \\ & x-x_0=a_0 \sin (\omega t+\alpha) \\ & \text { for } A=-B, C=2 B . \\ & \text { Amplitude } a_0=\sqrt{\left(\frac{B-A}{2}\right)^2+\left(\frac{C}{2}\right)^2}=\sqrt{2} B \end{aligned} $

From (i)

(A) For $\mathrm{A}=\mathrm{B}=0, x=\frac{\mathrm{C}}{2} \quad \sin 2 \omega t(\mathrm{SHM})$

(B) For $\mathrm{A}=-\mathrm{B}, \mathrm{C}=2 \mathrm{~B}$

$ \begin{aligned} & x=\frac{-\mathrm{B}+\mathrm{B}}{2}+\left(+\frac{\mathrm{B}}{2}+\frac{\mathrm{B}}{2}\right) \cos 2 \omega t+\frac{2 \mathrm{~B}}{2} \sin 2 \omega t \\ & x=\cos 2 \omega t+\mathrm{B} \sin 2 \omega t \\ & x=\sqrt{2} \mathrm{~B} \sin \left(2 \omega t+\frac{\pi}{4}\right)(\mathrm{SHM}) \end{aligned} $

Resultant amplitude $=\sqrt{2}$

(C) For $\mathrm{A}=\mathrm{B}, \mathrm{C}=0$,

$ \begin{aligned} x & =\frac{A+A}{2}+\left(\frac{A}{2}-\frac{A}{2}\right) \cos 2 \omega t+0 \\ & =\frac{A+A}{2}=A(\text { no SHM }) \end{aligned} $

$ \begin{aligned} & \text { (D) } \mathrm{A}=\mathrm{B}, \mathrm{C}=2 \mathrm{~B} \\ & \begin{aligned} x=\frac{\mathrm{A}+\mathrm{A}}{2}+\left(\frac{\mathrm{A}}{2}-\frac{\mathrm{A}}{2}\right) \cos 2 \omega t+\frac{2 \mathrm{~B}}{2} \sin 2 \omega t & \\ =\mathrm{A}+\mathrm{B} \sin 2 \omega t(\mathrm{OR}) \\ x=\mathrm{B}+\mathrm{B} \sin 2 \omega t(\mathrm{SHM}), \text { Amplitude }=|\mathrm{B}| \end{aligned} \end{aligned} $