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
$$\begin{aligned} N-m g & =m a=m \omega^2 A \\ \text{or}\quad N & =m g+m \omega^2 A \\ & =m\left[g+\omega^2 A\right] \\ \text{Putting the data}\quad N & =50\left[9.8+(4 \pi)^2 \times 5 \times 10^{-2}\right] \\ & =50\left[9.8+(12.56)^2 \times 5 \times 10^{-2}\right] \\ & =50[9.8+7.88] \\ & =50 \times 17.68=884 \mathrm{~N} \end{aligned}$$
$$ \begin{aligned} &\text { Now, the machine reads the normal reaction. It is clear that }\\ &\begin{aligned} & \text { maximum weight }=884 \mathrm{~N} \quad \text{(at lowest point)}\\ & \text { minimum weight }=95.5 \mathrm{~N} \quad \text{(at top point)} \end{aligned} \end{aligned}$$
A body of mass $m$ is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand, so that the spring is neither stretched nor compressed. Suddenly, the support of the hand is removed. The lowest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand.
(a) What is the amplitude of oscillation?
(b) Find the frequency of oscillation?
When the support of the hand is removed, the body oscillates about a mean position.
Suppose $x$ is the maximum extension in the spring when it reaches the lowest point in oscillation.
Loss in PE of the block $=m g x\quad \text{... (i)}$
where, $m=$ mass of the block
Gain in elastic potential energy of the spring
$$=\frac{1}{2} k x^2\quad\text{... (ii)}$$
As the two are equal, conserving the mechanical energy, we get,
$$m g x=\frac{1}{2} k x^2 \text { or } x=\frac{2 m g}{k}\quad \text{.... (iii)}$$
Now, the mean position of oscillation will be, when the block is balanced by the spring.
$$\begin{aligned} &\text { If } x^{\prime} \text { is the extension in that case, then }\\ &\begin{array}{lrl} & F & =+k x^{\prime} \\ \text { But } & F & =m g \\ \Rightarrow & m g & =+k x^{\prime} \\ \text { or } & x^{\prime} & =\frac{m g}{k}\quad \text{... (iv)} \end{array} \end{aligned}$$
Dividing Eq. (iii) by Eq. (iv),
$$\begin{aligned} \frac{x}{x^{\prime}} & =\frac{2 m g}{k} / \frac{m g}{k}=2 \\ x & =2 x^{\prime} \end{aligned}$$
But given $x=4 \mathrm{~cm}$ (maximum extension from the unstretched position)
$$\begin{array}{lr} \therefore & 2 x^{\prime}=4 \\ \therefore & x^{\prime}=\frac{4}{2}=2 \mathrm{~cm} \end{array}$$
But the displacement of mass from the mean position to the position when spring attains its natural length is equal to amplitude of the oscillation.
$$\therefore \quad A=x^{\prime}=2 \mathrm{~cm}$$
where, $A=$ amplitude of the motion.
(b) Time period of the oscillating system depends on mass spring constant given by
$$T=2 \pi \sqrt{\frac{m}{k}}$$
It does not depend on the amplitude.
But from Eq. (iii),
$$\begin{aligned} \frac{2 m g}{k} & =x \quad \text{(maximum extension)}\\ \frac{2 m g}{k} & =4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m} \\ \therefore\quad \frac{m}{k} & =\frac{4 \times 10^{-2}}{2 g}=\frac{2 \times 10^{-2}}{g} \\ \therefore \quad \frac{k}{m} & =\frac{g}{2 \times 10^{-2}} \\ \text{and}\quad v & =\text { frequency }=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \\ \therefore \quad v & =\frac{1}{2 \times 3.14} \sqrt{\frac{g}{2 \times 10^{-2}}} \\ & =\frac{1}{2 \times 3.14} \sqrt{\frac{4.9}{10^{-2}}}=\frac{1}{628} \times \sqrt{4.9 \times 100} \\ & =\frac{10}{628} \times 2.21=3.51 \mathrm{~Hz} . \end{aligned}$$
A cylindrical $\log$ of wood of height $h$ and area of cross-section $A$ floats in water. It is pressed and then released. Show that the log would execute SHM with a time period.
$$T=2 \pi \sqrt{\frac{m}{A \rho g}}$$
where, $m$ is mass of the body and $\rho$ is density of the liquid.
Consider the diagram,
Let the log be pressed and let the vertical displacement at the equilibrium position be $x_0$.
At equilibrium,
$$m g=\text { buoyant force }=\left(\rho A x_0\right) g \quad\left[\because m=V \rho=\left(A x_o\right) \rho\right]$$
When it is displaced by a further displacement $x$, the buoyant force is $A\left(x_0+x\right) \rho g$
$$\begin{aligned} \therefore \quad \text { Net restoring force } & =\text { Buoyant force }- \text { Weight } \\ & =A\left(x_0+x\right) \rho g-m g \end{aligned}$$
$$=(A \rho g) x \quad\left[\because m g=\rho A x_0 g\right]$$
As displacement $x$ is downward and restoring force is upward, we can write
$$\begin{aligned} F_{\text {restoring }} & =-(A \rho g) x \\ & =-k x\end{aligned}$$
where $$k=\text { constant }=A \rho g$$
So, the motion is SHM
($\because F \propto -x$)
$$\text { Now, } \quad \text { Acceleration } a=\frac{F_{\text {restoring }}}{m}=-\frac{k}{m} x$$
$$\begin{aligned} \text{Comparing with}\quad a & =-\omega^2 x \\ \Rightarrow \quad \omega^2 & =\frac{k}{m} \Rightarrow \omega=\sqrt{\frac{k}{m}} \\ \Rightarrow \quad \frac{2 \pi}{T} & =\sqrt{\frac{k}{m}} \Rightarrow T=2 \pi \sqrt{\frac{m}{k}} \\ \Rightarrow \quad T & =2 \pi \sqrt{\frac{m}{A \rho g}} \end{aligned}$$