Simple Harmonic Motion

The given situation is shown below.

$ \begin{aligned} &\text { Effective gravitational acceleration }\\ &\begin{aligned} & =\left(g-\frac{q E}{m}\right) \\ \therefore \quad T & =2 \pi \sqrt{\frac{l}{g-\frac{q E}{m}}} \end{aligned} \end{aligned} $

$ \text { The given situation is shown as } $

Let the angle be $\omega t$ when the KE is $75 \%$ of total energy.

Also, the moment of inertia of the bob of pendulum about the point of suspension

$ =I=m l^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, velocity at that instant

$ =l \omega \cos \omega t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, according to the question,

$ \frac{1}{2} m v^2=\frac{75}{100} \times \frac{1}{2} I \omega^2 $

where, $\left(\frac{1}{2} I \omega^2\right)$ is the rotational KE of the particle,

$ \begin{aligned} & \frac{1}{2} m v^2=\frac{75}{100} \times \frac{1}{2} \times m l^2 \times \omega^2 \text { [from Eq. (i)] } \\ & \Rightarrow \frac{1}{2} l^2 m \omega^2 \cos ^2 \omega t=\frac{3}{4} \times \frac{1}{2} \times m l^2 \omega^2 \end{aligned} $

[from Eq. (ii)]

$ \begin{aligned} & \Rightarrow \quad \cos ^2 \omega t=3 / 4 \\ & \Rightarrow \quad \cos \omega t= \pm \frac{\sqrt{3}}{2} \Rightarrow \omega t=\frac{\pi}{6} \end{aligned} $

(keeping only positive solution)

$ t=\frac{\pi}{6 \omega}=\frac{\pi T}{6 \times 2 \pi} $

( $\therefore$ Time period, $T=\frac{2 \pi}{\omega}$ )

$ \begin{array}{ll} \Rightarrow & t=\frac{\pi \times 4}{12 \pi} \,\,(given, T=4 \mathrm{~s} )\\ \Rightarrow & t=\frac{1}{3} \mathrm{~s} \end{array} $

Given, instantaneous displacement of particle executing SHM,

$\begin{aligned} x & =A \sin ^2(\omega t-4) \\ \Rightarrow x & =A\left[\frac{1-\cos 2(\omega t-4)}{2}\right] \\ & \left(\text { Since, } \sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\right) \end{aligned}$

$\Rightarrow x=\frac{A}{2}[1-\cos (2 \omega t-8)]$

Here, angular velocity of particle

$\begin{aligned} & \omega^{\prime}=2 \omega \\ & \Rightarrow \quad \frac{2 \pi}{T^{\prime}}=2 \times \frac{2 \pi}{T} \Rightarrow T^{\prime}=\frac{T}{2} \\ & =\frac{\frac{2 \pi}{\omega}}{2}=\frac{\pi}{\omega} \\ & \left(\therefore T=\frac{2 \pi}{\omega}\right) \\ \end{aligned}$

The mechanical energy lost in each cycle is proportional to the square of the amplitude $ A $. Therefore, the fractional change in energy can be expressed as:

$\begin{aligned} \frac{\Delta E}{E} \simeq \frac{dE}{E} & = \frac{dA^2}{A^2} \\ & = \frac{2AdA}{A^2} = 2 \frac{dA}{A} \\ & = 2 \times 1.5\% = 3\% \end{aligned}$

Thus, the mechanical energy of the oscillator lost in each cycle is 3%.

According to first condition,

In S. H. M, at displacement $x$

Potential energy $=9 \mathrm{~J}$

-i.e, $\frac{1}{2} k x^2=9 \quad \text{... (i)}$

Again, at displacement $y$, Potential energy

$\frac{1}{2} k y^2=16 \quad \text{... (ii)}$

Dividing Eq. (ii) by Eq. (i), we get

$\begin{aligned} \frac{\frac{1}{2} k y^2}{\frac{1}{2} k x^2} & =\frac{16}{9} \\ \Rightarrow \quad\left(\frac{y}{x}\right)^2 & =\left(\frac{4}{3}\right)^2 \\ \frac{y}{x} & =\frac{4}{3} \quad \text{... (iii)} \end{aligned}$

$\because$ Potential energy of body at displacement $(x+y)$

$\begin{aligned} \mathrm{PE} & =\frac{1}{2} k(x+y)^2 \\ & =\frac{9}{x^2}(x+y)^2 \quad \text{[From Eq. (i)]}\\ & =9 \frac{(x+y)^2}{x^2}=9\left(1+\frac{y}{x}\right)^2 \\ & =9\left(1+\frac{4}{3}\right)^2 \quad \text{[From Eq. (iii)]}\\ & =49 \mathrm{~J} \end{aligned}$

If the hydrometer is in equilibrium, the weight is balanced by the buoyancy force on it by the fluid. When hydrometer liquid is pressed by $x$ distance the net unbalanced force in the upward direction is given by $F=-$(Area $\times$ Density $\times$ Gravity $\times$ Distance moved by fluid)

Where, $-v e$ sign shows that the direction of force will be opposite to the direction of motion.

Thus, $m \frac{d^2 x}{d t^2}=-\pi r^2 \rho g x$

Were, $r$ = radius of liquid column

$\rho=$ density of liquid

$x=$ distance moved by liquid

Now, $\quad \frac{d^2 x}{d t^2}=\frac{-\pi r^2 \rho g}{m} x$

or $\quad \omega^2=\frac{\pi r^2 \rho g}{m}$ or $\omega=\sqrt{\frac{\pi r^2 \rho g}{m}}$

$\Rightarrow \quad T=2 \pi \sqrt{\frac{m}{\pi r^2 \rho g}}$

The velocity of the object becomes zero at maximum displacement of either side of the mean position in the SHM.

Time taken from one maximum to another maximum $=0.5 \mathrm{~s}$

Time period $=2 \times 0.5=1 \mathrm{~s}$

Angular velocity, $\omega=2 \pi v$

$=\frac{2 \pi}{T}=\frac{2 \pi}{l}=2 \pi \mathrm{~rad} / \mathrm{s}$

Let initially the cone be in equilibrium i.e. floating

$\begin{aligned} & \text { From the diagram, it is clear that } \\ & \text { Buoyant force }\left(F_b\right)=\text { weight of the cone. }\left(F_w\right) \\ & \qquad \rho_w \times\left(\frac{1}{3} \pi r^2 x\right) \times g=\rho_c \times\left(\frac{1}{3} \pi R^2 H\right) \times g \end{aligned}$

$\Rightarrow \quad 2 r^2 x=R^2 H \quad\left(\because \rho_c=\frac{\rho_w}{2}\right)$

$\Rightarrow \quad x=\frac{R^2 H}{2 r^2} \quad \text{... (i)}$

From the diagram,

$\begin{array}{ll} & \tan \theta=\frac{r}{x}=\frac{R}{H} \\ \Rightarrow \quad & x=\frac{r H}{R} \quad \text{... (ii)} \end{array}$

From Eqs. (i) and (ii)

$\begin{gathered} \frac{R^2 H}{2 r^2}=\frac{r H}{R} \\ \Rightarrow \quad \frac{R^3}{2}=r^3 \Rightarrow r=\left(\frac{1}{2}\right)^{1 / 3} R \end{gathered}$

Now, the cone. is depressed by small distance $\delta(<< H)$ and released it executes simple harmonic motion. In simple harmonic motion, the buoyancy force acts as restoring force which is given as

$F_b=\text { density } \times \text { volume } \times \text { gravity }$

Since, the displacement is very small, the excess submerged shape of cone. can be considered as a cylinder whose volume is given by $\pi r^2 \delta$.

$\therefore \quad F_b=\rho_w \pi r^2 \delta g=\rho_w \pi\left[\left(\frac{1}{2}\right)^{1 / 3} R\right]^2 \delta g$

Thus, force is equal to restoring force, hence

$F_b=k \delta=m \omega^2 \delta$

$\begin{array}{ll} & \rho_w \pi\left(\frac{1}{4}\right)^{1 / 3} R^2 \delta g=\rho_c \frac{1}{3} \pi R^2 H \delta \omega^2 \\ & 2 \times 3 \times\left(\frac{1}{4}\right)^{1 / 3} \delta g=H \delta \omega^2 \quad\left(\because \rho_c=\frac{\rho_w}{2}\right) \\ \Rightarrow \quad & \frac{6\left(\frac{1}{4}\right)^{1 / 3} g}{H}=\omega^2 \\ \text { or } & \omega=\sqrt{\frac{6 g}{H}\left(\frac{1}{4}\right)^{1 / 3}} \Rightarrow 2 \pi f=\sqrt{\frac{6 g}{H}\left(\frac{1}{4}\right)^{1 / 3}} \\ \Rightarrow \quad & f=\frac{1}{2 \pi} \sqrt{\frac{6 g}{H}\left(\frac{1}{4}\right)^{1 / 3}}=\frac{1}{2 \pi} \sqrt{\frac{6 g}{H} \cdot \frac{1}{4^{1 / 3}}} \end{array}$

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} $

As we know that, for less damping, resonance peak is taller and sharper.

Let, wave displacement be x.

As we know that,

Potential energy, $U=\frac{1}{2}kx^2$

where, k is wave constant.

Potential energy will be maximum, when displacement is maximum.

$\Rightarrow x=\pm~A$

Let,

Time period $=T$

Length of simple pendulum $=l$

Acceleration due to gravity $=g$

$\because \quad T=2 \pi \sqrt{l / g}$ .... (i)

When sphere is fully filled with water, centre of mass lies at centre.

Case I For, water level below centre of sphere

$l_1 > l_0$ ...... (ii)

Case II For empty sphere,

$l=l_0 < l_1$ ..... (iii)

$\therefore$ From Eqs. (i), (ii) and (iii)

In case I, time period increase.

In case II, time period decrease.

Given, mass of block, $m=1 \mathrm{~kg}$

Spring constant, $k=100 \mathrm{~N} / \mathrm{m}$

In case I, extension in spring, $\Delta x_1=10 \mathrm{~cm}$

$=10 \times 10^{-2} \mathrm{~m}$

Initial speed, $u=0 \mathrm{~ms}^{-1}$

In case II, extension in spring $\Delta x_2=5 \mathrm{~cm}$

$=5 \times 10^{-2} \mathrm{~m}$

Let, kinetic energy and potential energy be (KE) and (PE)

As we know that,

$\begin{aligned} & \mathrm{KE}=\frac{1}{2} k\left(\Delta x_1^2-\Delta x_2^2\right) \\ &=\frac{1}{2} \times 100\left(10^2-5^2\right) \times 10^{-4} \\ &=50 \times 75 \times 10^{-4}=0.375 \mathrm{~J} \\ & \text { and } \mathrm{PE}=\frac{1}{2} k \Delta x_2^2 \\ &=\frac{1}{2} \times 100 \times\left(5 \times 10^{-2}\right)^2=50 \times 25 \times 10^{-4}=0.125 \mathrm{~J} \end{aligned}$