Oscillations and Periodic Motion

A seconds pendulum is a pendulum whose period is precisely two seconds

The fractional change in time period is given by delta T / T = 1/2 * alpha * delta theta. With alpha = 12*10^-6 and delta theta = 20 degrees, the fractional change is 1.2*10^-4. Multiplying by the number of seconds in a day (86400), we get approximately 10.36 seconds.

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ .. (1)

We get $1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ $\Rightarrow\, L\, =\, \displaystyle \frac{g}{4\, \pi^2}\, =\, \displaystyle \frac{9.8}{4\, \times\, 3.14\, \times\, 3.14}m\, \Rightarrow\, L\, =\, \displaystyle \frac{9.8}{39.44}m\, =\, 0.2484\, m\, =\, 0.25\, m$ (approx.)

Two pendulums of length $121cm$ and $100cm$.

As block is floating it's weight should be equal to buoyancy force 
$\rho _{wood} V _{cylinder} g=\rho _{liquid} V _{displaced} g$
$V _{displaced} =\frac{\rho _{wood} V _{cylinder} }{\rho _{liquid}} $
$V _{displaced} =\frac{\rho _{wood} V _{cylinder} }{\rho _{liquid}} $...(i)
After displacing by small distance x, the net force on cylinder will be
$F=Buoyancy-w=\rho _{liquid}( V _{displaced} +A _{cylinder}\Delta x) g- \rho _{wood} V _{cylinder} g$
$F=\rho _{liquid}  V _{displaced} g+\rho _{liquid} A _{cylinder}\Delta x g-\rho _{wood} V _{wood} g$ the net force on cylinder becomes 
from equation (i) $\rho _{wood} V _{cylinder} g=\rho _{liquid} V _{displaced} g$
$F=\rho _{liquid} A _{cylinder}\Delta x g$
$ma=\rho _{liquid} A _{cylinder}\Delta x g$
$a=\frac{\rho _{liquid} A _{cylinder}}{m}\Delta x $
$a = \omega^2 \Delta x $
$\omega^2 =\frac{\rho _{liquid} A _{cylinder}}{\rho _{cylinder} V _{cylinder}}$
$\omega^2 =\frac{\rho _{liquid} A _{cylinder}}{\rho _{wood} A _{cylinder} h _{cylinder}}$
$\omega^2 =\frac{\rho _{liquid}}{\rho _{wood} h _{cylinder}}$
$\omega^2 =\frac{900}{650\times .54 }$
This should be equal to angular frequency of simple pendulum
$ \omega=\sqrt{\frac{g}{l}}$
$\sqrt{\frac{900}{650\times .54}}=\sqrt{\frac{g}{l}}$
$l=g\frac{650\times .54}{900}$
$l=.06\times 65$
$=39 cm$

In case of vibration magnetometer when a magnet is cut n equal parts by cutting normal to its length. Then the time period of each part of magnet will be 
$T'=\frac{T}{n}$       ...(i)   (here, $T=4s, n = 2$)
Now, Putting the given values in Eq. (i), we get
$T'=\frac{4}{2}=2s$

Time period of vibration: $T \propto \sqrt{\dfrac{I}{M}}$ where $I$ is the moment of inertia and $M$ is the magnetic moment
$ \therefore$ $\dfrac{T _1}{T _2}= \sqrt{\dfrac{I _1M _2}{I _2M _1}}$
When the magnet is cut into 4 pieces parallel to its length, magnetic moment remains same since the breadth is very small..
$\therefore M _1= M _2$
Moment of inertial also  does not change.
$I _2=I _1$
$ \therefore \dfrac{T _1}{T _2}=\sqrt{\dfrac{I _1M _2}{I _2\times M _1}}=1$
$ \therefore T _2 = T _1=4$

Time period of magnet, $T=2\pi \sqrt { \dfrac { I }{ MB }  } $
When magnet is cut parallel to its length into four equal pieces.
Then new
magnetic moment, ${ M }^{ \prime  }=\dfrac { M }{ 4 } $
New moment of inertia, ${ I }^{ \prime  }=\dfrac { I }{ 4 } $
$\therefore $ New time period, ${ T }^{ \prime  }=2\pi \sqrt { \dfrac { { I }^{ \prime  } }{ { M }^{ \prime  }{ B }^{ \prime  } }  } $
$\Rightarrow \quad T={ T }^{ \prime  }=4s$

A pendulum watch relies on gravity to oscillate. Inside an orbiting satellite, objects are in a state of free fall, meaning the effective gravitational acceleration (g) is zero, causing the pendulum to stop oscillating.

Sound is heard only if the body vibrates with a frequency of at least 20 Hz. Frequency of vibration of the pendulum is less than 20 Hz. Hence no sound is heard when the pendulum vibrates or oscillates in air.

The low frequency vibration of infrasonic wave is less than $20\,Hz$. The pendulum vibrates at this low frequency with a time period of $1$ second.

As temperature increases in summer, the metal pendulum rod undergoes thermal expansion, increasing its length. Since the period of a pendulum is proportional to the square root of its length, a longer pendulum results in a longer time period, causing the clock to run slow.

The vibrations of the pendulum die after some time , it is because of the frictional force of air which opposes the motion of bob .So air resistance is responsible for the loss in mechanical energy .