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

$$\begin{aligned} N-m g & =m a=m \omega^2 A \\ \text{or}\quad N & =m g+m \omega^2 A \\ & =m\left[g+\omega^2 A\right] \\ \text{Putting the data}\quad N & =50\left[9.8+(4 \pi)^2 \times 5 \times 10^{-2}\right] \\ & =50\left[9.8+(12.56)^2 \times 5 \times 10^{-2}\right] \\ & =50[9.8+7.88] \\ & =50 \times 17.68=884 \mathrm{~N} \end{aligned}$$

$$ \begin{aligned} &\text { Now, the machine reads the normal reaction. It is clear that }\\ &\begin{aligned} & \text { maximum weight }=884 \mathrm{~N} \quad \text{(at lowest point)}\\ & \text { minimum weight }=95.5 \mathrm{~N} \quad \text{(at top point)} \end{aligned} \end{aligned}$$

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),

$$\begin{aligned} \frac{2 m g}{k} & =x \quad \text{(maximum extension)}\\ \frac{2 m g}{k} & =4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m} \\ \therefore\quad \frac{m}{k} & =\frac{4 \times 10^{-2}}{2 g}=\frac{2 \times 10^{-2}}{g} \\ \therefore \quad \frac{k}{m} & =\frac{g}{2 \times 10^{-2}} \\ \text{and}\quad v & =\text { frequency }=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \\ \therefore \quad v & =\frac{1}{2 \times 3.14} \sqrt{\frac{g}{2 \times 10^{-2}}} \\ & =\frac{1}{2 \times 3.14} \sqrt{\frac{4.9}{10^{-2}}}=\frac{1}{628} \times \sqrt{4.9 \times 100} \\ & =\frac{10}{628} \times 2.21=3.51 \mathrm{~Hz} . \end{aligned}$$

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

($\because F \propto -x$)

$$\text { Now, } \quad \text { Acceleration } a=\frac{F_{\text {restoring }}}{m}=-\frac{k}{m} x$$

$$\begin{aligned} \text{Comparing with}\quad a & =-\omega^2 x \\ \Rightarrow \quad \omega^2 & =\frac{k}{m} \Rightarrow \omega=\sqrt{\frac{k}{m}} \\ \Rightarrow \quad \frac{2 \pi}{T} & =\sqrt{\frac{k}{m}} \Rightarrow T=2 \pi \sqrt{\frac{m}{k}} \\ \Rightarrow \quad T & =2 \pi \sqrt{\frac{m}{A \rho g}} \end{aligned}$$

Consider the diagram shown below

Let us consider an infinitesimal liquid column of length $d x$ at a height $x$ from the horizontal line.

If $\rho=$ density of the liquid

$$A=\text { cross-sectional area of } V \text {-tube }$$

PE of element $d x$ will be given as

$$P E=d m g x=(A p d x) g x \quad[\because d m=p V=p A d x]$$

where $$\quad A p d x=d m=\text { mass of element } d x$$

$\therefore$ Total PE of the left column

$$\begin{aligned} & =\int_0^{h_1} A \rho g x d x \\ & =A \rho g \int_0^{h_1} x d x \\ & =A \rho g\left|\frac{x^2}{2}\right|_0^{h_1}=A \rho g \frac{h_1^2}{2} \end{aligned}$$

$$\begin{aligned} &\text { But, }\\ &h_1=l \sin 45^{\circ}\\ &\therefore \quad P E=\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ} \quad \text{... (i)} \end{aligned}$$

In a similar way,

$$ \begin{aligned} \text { PE of right column } & =\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ} \\ \text { Total PE } & =\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ}+\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ} \quad \text{... (ii)}\\ & =2 \times \frac{1}{2} A \rho g l^2\left(\frac{1}{\sqrt{2}}\right)^2=\frac{A \rho g l^2}{2}\quad \text{... (iii)} \end{aligned}$$

If due to pressure difference created y element of left side moves on the right side, then

liquid present in the left arm $=l-y$

liquid present in the right arm $=l+y$

$$\begin{array}{ll} \therefore \quad & \text { Total PE }=A \rho g(l-y)^2 \sin ^2 45^{\circ}+A \rho g(l+y)^2 \sin ^2 45^{\circ} \\ & \text { Changes in } \mathrm{PE}=(\mathrm{PE})_{\text {tinal }}-(\mathrm{PE})_{\text {initial }} \end{array}$$

$$\begin{aligned} \text{or}\quad \Delta \mathrm{PE} & =\frac{A \rho g}{2}\left[(l-y)^2+(l+y)^2-l^2\right] \\ & =\frac{A \rho g}{2}\left[l^2+y^2-2 l y+l^2+y^2+2 l y-l^2\right] \\ & =\frac{A \rho g}{2}\left[2\left(l^2+y^2\right)\right] \\ & =A \rho g\left(l^2+y^2\right)\quad \text{... (iv)} \end{aligned}$$

If $v$ is the change in velocity of the total liquid column, then change in KE

$$\begin{aligned} \Delta \mathrm{KE} & =\frac{1}{2} m v^2 \\ \text{But}\quad m & =A \rho(2 l) \\ \therefore \quad \Delta \mathrm{KE} & =\frac{1}{2} A \rho 2 l v^2=A \rho l v^2\quad \text{... (v)} \end{aligned}$$

From Eqs. (iv) and (v),

$$\Delta \mathrm{PE}+\Delta \mathrm{KE}=A \rho g\left(l^2+y^2\right)+A \rho l v^2\quad \text{... (vi)}$$

System being conservative.

$\therefore$ Change in total energy $=0$

From Eq. (vi), $$\quad A \rho g\left(l^2+y^2\right)+A \rho l v^2=0$$

Differentiating both sides with respect to time $(t)$, we get

$$\begin{aligned} & A \rho g\left[0+2 y \frac{d y}{d t}\right]+A \rho l(2 v) \frac{d v}{d t}=0 \\ \text{But,}\quad & \frac{d y}{d t}=v \text { and } \frac{d v}{d t}=a \quad \text{[acceleration]} \end{aligned}$$

$$\begin{array}{l} \Rightarrow \quad A \rho g(2 y v)+A \rho l(2 v) a=0 \\ \Rightarrow \quad (g y+l a) 2 A \rho v=0 \\ 2 A \rho v=\text { constant and } 2 A \rho v \neq 0 \end{array}$$

$$ \begin{aligned} \therefore \quad l a+g y & =0 \\ a+\left(\frac{g}{l}\right) y & =0 \\ \text{or}\quad \frac{d^2 y}{d t^2}+\left(\frac{g}{l}\right) y & =0 \end{aligned}$$

$$\begin{aligned} &\text { This is the standard differential equation for SHM of the form }\\ &\begin{aligned} \frac{d^2 y}{d t^2}+\omega^2 y & =0 \\ \therefore\quad \omega & =\sqrt{\frac{g}{l}} \\ \therefore\quad T=\frac{2 \pi}{\omega} & =2 \pi \sqrt{\frac{l}{g}} \end{aligned} \end{aligned}$$

Consider the situation shown in the diagram.

The gravitational force on the particle at a distance $r$ from the centre of the earth arises entirely from that portion of matter of the earth in shells internal to the position of the particle. The external shells exert no force on the particle.

More clearly, let $g^{\prime}$ be the acceleration at $P$.

So, $$g^{\prime}=g\left(1-\frac{d}{R}\right)=g\left(\frac{R-d}{R}\right)$$

$$\begin{aligned} &\begin{aligned} &\text { From figure, }\quad R-d =y \\ \Rightarrow \quad g^{\prime} & =g \frac{y}{R} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { Force on body at } p \text {, }\\ &F=-m g^{\prime}=\frac{-m g}{R} y\quad \text{... (i)} \end{aligned}$$

$$\begin{aligned} &\Rightarrow \quad F \propto-y \quad \text { [where, } y \text { is distance from the centre] }\\ &\text { So, motion is SHM. } \end{aligned}$$

For time period, we can write Eq. (i)

As $$m a=-\frac{M g}{R} y \Rightarrow a=-\frac{g}{R} y$$

Comparing with $a=-\omega^2 y$

$$\begin{aligned} \omega^2 & =\frac{g}{R} \\ \Rightarrow \quad\left(\frac{2 \pi}{T}\right) & =\frac{g}{R} \Rightarrow T=2 \pi \sqrt{\frac{R}{g}} \end{aligned}$$