Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.
Let us assume that the required displacement be $x$.
$\therefore$ Potential energy of the simple harmonic oscillator $=\frac{1}{2} k x^2$
$$\begin{aligned} \text{where,}\quad k & =\text { force constant }=m \omega^2 \\ \therefore \quad \mathrm{PE} & =\frac{1}{2} m \omega^2 x^2\quad \text{... (i)} \end{aligned}$$
Maximum energy of the oscillator
$$ \begin{aligned} &\mathrm{TE}=\frac{1}{2} m \omega^2 A^2 \quad\left[\because x_{\max }=A\right] \quad\text { ....(ii) } \end{aligned}$$
$$\begin{aligned} \text{where,}\quad A & =\text { amplitude of motion } \\ \text{Given,}\quad\mathrm{PE} & =\frac{1}{2} \mathrm{TE} \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad & \frac{1}{2} m \omega^2 x^2 =\frac{1}{2}\left[\frac{1}{2} m \omega^2 A^2\right] \\ \Rightarrow \quad & x^2 =\frac{A^2}{2} \\ \text { or } \quad & x =\sqrt{\frac{A^2}{2}}= \pm \frac{A}{\sqrt{2}} \end{aligned}$$
Sign $$\pm$$ indicates either side of mean position.
A body of mass $m$ is situated in a potential field $U(x)=U_0(1-\cos \alpha x)$ when, $U_0$ and $\alpha$ are constants. Find the time period of small oscillations.
Given potential energy associated with the field
$$U(x)=U_0(1-\cos \alpha x)\quad \text{... (i)}$$
Now, force $F=-\frac{d U(x)}{d x}$
$$\left[\because \text { for conservatine force } f, \text { we can write } f=\frac{-d u}{d x}\right]$$
$$\begin{aligned} &\text { [We have assumed the field to be conservative] }\\ &\begin{aligned} & F=-\frac{d}{d x}\left(U_0-U_0 \cos \alpha x\right)=-U_0 \alpha \sin \alpha x \\ & F=-U_0 \alpha^2 x\quad \text{... (ii)} \end{aligned} \end{aligned}$$
$$[\because \text { for small oscillations } \alpha x \text { is small, } \sin \alpha x \approx \alpha x \text { ] }$$
$$\Rightarrow \quad F \propto(-x)$$
As, $U_0, \alpha$ being constant.
$\therefore$ Motion is SHM for small oscillations.
Standard equation for SHM
$$F=-m \omega^2 x\quad \text{... (iii)}$$
Comparing Eqs. (ii) and (iii), we get
$$\begin{aligned} m \omega^2 & =U_0 \alpha^2 \\ \omega^2 & =\frac{U_0 \alpha^2}{m} \text { or } \omega=\sqrt{\frac{U_0 \alpha^2}{m}} \\ \therefore \quad \text { Time period } T & =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{U_0 \alpha^2}} \end{aligned}$$
A mass of 2 kg is attached to the spring of spring constant $50 \mathrm{Nm}^{-1}$. The block is pulled to a distance of 5 cm from its equilibrium position at $x=0$ on a horizontal frictionless surface from rest at $t=0$. Write the expression for its displacement at anytime $t$.
Consider the diagram of the spring block system. It is a SHM with amplitude of 5 cm about the mean position shown.
Given, $\quad$ spring constant $k=50 \mathrm{~N} / \mathrm{m}$
$m=$ mass attached $=2 \mathrm{~kg}$
$\therefore \quad$ Angular frequency $\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5 \mathrm{rad} / \mathrm{s}$
Assuming the displacement function
$$y(t)=A \sin (\omega t+\phi)$$
where, $\phi=$ initial phase
But given at $t=0, y(t)=+A$
$$y(0)=+A=A \sin (\omega \times 0+\phi)$$
or $$\quad \sin \phi=1 \Rightarrow \phi=\frac{\pi}{2}$$
$\therefore$ The desired equation is $\quad y(t)=A \sin \left(\omega t+\frac{\pi}{2}\right)=A \cos \omega t$
Putting $A=5 \mathrm{~cm}, \omega=5 \mathrm{rad} / \mathrm{s}$
we get, $y(t)=5 \sin 5 t$
where, $t$ is in second and $y$ is in centimetre.
Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^{\circ}$ to the right with the vertical, the other pendulum makes an angle of $1^{\circ}$ to the left of the vertical. What is the phase difference between the pendulums?
Consider the situations shown in the diagram (i) and (ii)
Assuming the two pendulums follow the following functions of their angular displacements
$$\begin{array}{ll} & \theta_1=\theta_0 \sin \left(\omega t+\phi_1\right) \quad \text{... (i)}\\ \text { and } & \theta_2=\theta_0 \sin \left(\omega t+\phi_2\right)\quad \text{... (ii)} \end{array}$$
As it is given that amplitude and time period being equal but phases being different.
Now, for first pendulum at any time $t$
$\theta_1=+\theta_0\quad$ [Right extreme]
From Eq. (i), we get
$$\begin{aligned} \Rightarrow\quad \theta_0 & =\theta_0 \sin \left(\omega t+\phi_1\right) \text { or } 1=\sin \left(\omega t+\phi_1\right) \\ \Rightarrow\quad \sin \frac{\pi}{2} & =\sin \left(\omega t+\phi_1\right) \\ \text{or}\quad \left(\omega t+\phi_1\right) & =\frac{\pi}{2}\quad \text{... (iii)} \end{aligned}$$
Similarly, at the same instant t for pendulum second, we have
$$\theta_2=-\frac{\theta_0}{2}$$
where $\theta_0=2^{\circ}$ is the angular amplitude of first pendulum. For the second pendulum, the angular displacement is one degree ,therefore $\theta_2=\frac{\theta_0}{2}$ and negative sign is taken to show for being left to mean position.
From Eq. (ii), then
$$-\frac{\theta_0}{2}=\theta_0 \sin \left(\omega t+\phi_2\right)$$
$$\begin{aligned} \Rightarrow\quad\sin \left(\omega t+\phi_2\right) & =-\frac{1}{2} \Rightarrow\left(\omega t+\phi_2\right)=-\frac{\pi}{6} \text { or } \frac{7 \pi}{6} \\ \text{or}\quad \left(\omega t+\phi_2\right) & =-\frac{\pi}{6} \text { or } \frac{7 \pi}{6}\quad \text{.... (iv)} \end{aligned}$$
$$\begin{aligned} &\text { From Eqs. (iv) and (iii), the difference in phases }\\ &\begin{gathered} \left(\omega t+\phi_2\right)-\left(\omega t+\phi_1\right)=\frac{7 \pi}{6}-\frac{\pi}{2}=\frac{7 \pi-3 \pi}{6}=\frac{4 \pi}{6} \\ \text { or }\quad \left(\phi_2-\phi_1\right)=\frac{4 \pi}{6}=\frac{2 \pi}{3}=120^{\circ} \end{gathered} \end{aligned}$$
A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of $2.0 \mathrm{~s}^{-1}$ and an amplitude 5.0 cm . A weighing machine on the platform gives the persons weight against time.
(a) Will there be any change in weight of the body, during the oscillation?
(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?
In this case acceleration is variable. In accelerated motion, weight of body depends on the magnitude and direction of acceleration for upward or downward motion.
(a) Hence, the weight of the body changes during oscillations
(b) Considering the situation in two extreme positions, as their acceleration is maximum in magnitude.
we have,
$$m g-N=m a$$
$\because$ At the highest point, the platform is accelerating downward.
$$\begin{aligned} \Rightarrow \quad N & =m g-m a \\ \text{But}\quad a & =\omega^2 A \quad \text{[in magnitude]}\\ \therefore \quad N & =m g-m \omega^2 A \\ \text{where,}\quad A & =\text { amplitude of motion. } \\ \text{Given,}\quad m & =50 \mathrm{~kg}, \text { frequency } v=2 \mathrm{~s}^{-1} \\ \therefore\quad \omega & =2 \pi v=4 \pi \mathrm{rad} / \mathrm{s} \\ A & =5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m} \\ \therefore \quad N & =50 \times 9.8-50 \times(4 \pi)^2 \times 5 \times 10^{-2} \\ & =50\left[9.8-16 \pi^2 \times 5 \times 10^{-2}\right] \\ & =50[9.8-7.89] \\ & =50 \times 1.91 \\ & =95.5 \mathrm{~N} \end{aligned}$$
When the platform is at the lowest position of its oscillation,
It is accelerating towards mean position that is vertically upwards. Writing equation of motion
$$\begin{aligned} N-m g & =m a=m \omega^2 A \\ \text{or}\quad N & =m g+m \omega^2 A \\ & =m\left[g+\omega^2 A\right] \\ \text{Putting the data}\quad N & =50\left[9.8+(4 \pi)^2 \times 5 \times 10^{-2}\right] \\ & =50\left[9.8+(12.56)^2 \times 5 \times 10^{-2}\right] \\ & =50[9.8+7.88] \\ & =50 \times 17.68=884 \mathrm{~N} \end{aligned}$$
$$ \begin{aligned} &\text { Now, the machine reads the normal reaction. It is clear that }\\ &\begin{aligned} & \text { maximum weight }=884 \mathrm{~N} \quad \text{(at lowest point)}\\ & \text { minimum weight }=95.5 \mathrm{~N} \quad \text{(at top point)} \end{aligned} \end{aligned}$$