Simple Harmonic Motion Concept Page - 9

Formula

Angular Displacement and Angular Velocity of a Physical Pendulum

Angular Displacement of Physical pendulum:
θ=θosin(wt+ϕ)$\theta ={\theta }_{o}sin(wt+\varphi )$
Angular Velocity of physical pendulum:
ω=θowsin(wt+ϕ)$\omega ={\theta }_{o}wsin(wt+\varphi )$

Example

Comment on the tension in the string of a simple pendulum at mean and extreme position of motion

The Mean Position is the Position that is moderate between two other extreme positions. It is the Position of the Bob when the freely suspended Pendulum is at rest.
Example: Determine the Tension in the String of a Simple Pendulum at Mean and Extreme Positions.

Solution:
Let  m$m$  and   L$L$  be the mass of the bob and length of the string respectively.Using circular motion equation:     mv2L=T−mg${\displaystyle \frac{m{v}^2}{L}}=T-mg$
∴T=mv2L+mg$\therefore T={\displaystyle \frac{m{v}^2}{L}}+mg$
As we know that  velocity  (v)$(v)$ is maximum at the mean position, thus
according to the equation  tension  (T)$(T)$ is maximum at the mean
position.Also  velocity  (v)$(v)$ is zero at extreme position, thus  T$T$  is minimum at extreme position.

Example

Write the equations of displacement and velocity of a simple pendulum

Example: A simple pendulum performs SHM about x=0$x=0$ with amplitude A$A$ and time period T$T$. What will be the speed of the pendulum at x=A/2$x=A/2$ ?

Solution:

v=ωA2−x2−−−−−−−√v=ωA2−(A2)2−−−−−−−−−−√=(2πT)(3√A2)=πA3√T$$\begin{gathered} v=\omega \sqrt{A^2-x^2} \\ v=\omega \sqrt{A^2-{\left({\displaystyle \frac{A}{2}}\right)}^2}=\left({\displaystyle \frac{2\pi }{T}}\right)\left({\displaystyle \frac{\sqrt{3}A}{2}}\right)=\pi A{\displaystyle \frac{\sqrt{3}}{T}} \end{gathered}$$

Example

Problem on time period of simple pendulum

Example: The bob of a simple pendulum executes simple harmonic motion in water with a period t$\mathrm{t}$, while the period of oscillation of the bob is t0${\mathrm{t}}_0$ in air. Neglecting frictional force of water and given that the density of the bob is (4/3)×1000 kg/m3$(4/3)\times 1000\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$. What relationship between t$\mathrm{t}$ and t0${\mathrm{t}}_0$ is true? 

Solution:
Let  V$V$ be the volume of the bob.

Thus mass of the bob       M=ρbV=43×1000V kg$M={\rho }_{b}V={\displaystyle \frac{4}{3}}\times 1000Vkg$Acceleration due to gravity in air is  g.$g.$   

Effective weight of the bob in water    :   Mg′=Mg−B$M{g}^{\prime }=Mg-B$       where  buoyant force  B=(Vρwater)g$B=(V{\rho }_{water})g$
43×1000Vg′=43×1000Vg−V×1000g${\displaystyle \frac{4}{3}}\times 1000V{g}^{\prime }={\displaystyle \frac{4}{3}}\times 1000Vg-V\times 1000g$  (as ρwater=1000kg/m3${\rho }_{water}=1000kg/m^3$)
43×1000Vg′=13×1000Vg${\displaystyle \frac{4}{3}}\times 1000V{g}^{\prime }={\displaystyle \frac{1}{3}}\times 1000Vg$  ⟹g′=g4$⟹g^{\prime }={\displaystyle \frac{g}{4}}$  Now  time period of the bob in air           to=2πlg−−√$t_o=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$Time period of the bob in water          t=2πlg′−−√$t=2\pi \sqrt{{\displaystyle \frac{l}{g^{\prime }}}}$ =2π4lg−−√$=2\pi \sqrt{{\displaystyle \frac{4l}{g}}}$   =2×2πlg−−√$=2\times 2\pi \sqrt{{\displaystyle \frac{l}{g}}}$⟹t=2to$⟹t=2t_o$

Example

Write the equations of angular velocity of a simple pendulum

Example: The angular frequency of a simple pendulum is ω$\omega$ rad/sec. Now, if the length is made one-fourth of the original length, then what is the angular frequency?

Solution:
The angular frequency of simple pendulum is ω=gl−−√$\omega =\sqrt{{\displaystyle \frac{g}{l}}}$
The new angular frequency when length is reduced to one fourth is
ωnew=gl4−−−⎷=4gl−−−√=2ω${\omega }_{new}=\sqrt{{\displaystyle \frac{g}{{\displaystyle \frac{l}{4}}}}}=\sqrt{{\displaystyle \frac{4g}{l}}}=2\omega$

Example

Describe a torsional pendulum and write torque equation for a torsional pendulum

A torsional pendulum consists of a disk-like mass suspended from a thin rod or wire. When the mass is twisted about the axis of the wire, the wire exerts a torque on the mass, tending to rotate it back to its original position. Upon giving a twist of θ$\theta$ the restoring torque produced is:

τ=−kθ$\tau =-k\theta$

Definition

Write angular displacement and angular velocity of a torsional pendulum

Angular Displacement of torsional pendulum:
θ=θosin(wt+ϕ)$\theta ={\theta }_{o}sin(wt+\varphi )$
Angular Velocity of torsional pendulum:
ω=θowsin(wt+ϕ)$\omega ={\theta }_{o}wsin(wt+\varphi )$

Formula

Time Period of Torsional Pendulum

Iα=−kθ$I\alpha =-k\theta$
or, Id2θdt2=−kθ$I{\displaystyle \frac{d^2\theta }{d{t}^2}}=-k\theta$
General Solution:
θ=θosin(wt+ϕ)$\theta ={\theta }_{o}sin(wt+\varphi )$

And time period: T=2πI/k−−−√$T=2\pi \sqrt{I/k}$

Formula

Write the torque equation of a physical pendulum for small angles of deviation

Restoring torque produced in a physical pendulum for small angles of deviation:
τ=−kθ$\tau =-k\theta$Iα=−kθ$I\alpha =-k\theta$

Example

Simple pendulum in accelerated frame of reference

Example:
A system pendulum is oscillating in a lift.If the lift is going down with constant velocity,the time period of the simple pendulum is T1$T_1$.If the lift is going down with some retardation its time period is T2$T_2$. Find relation between T1$T_1$ and T2$T_2$.
Solution:
Time period of pendulum with constant velocity is given by
T1=2πlg−−√$T_1=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$
while the time period of pendulum in a lift moving downwards with constant retardation is given by
T2=2πlgeff−−−−√=2πlg−(−a)−−−−−−−−√=2πlg+a−−−−−√$T_2=2\pi \sqrt{{\displaystyle \frac{l}{g_{eff}}}}=2\pi \sqrt{{\displaystyle \frac{l}{g-(-a)}}}=2\pi \sqrt{{\displaystyle \frac{l}{g+a}}}$ 
its clear from above two equations that T1>T2$T_1>T_2$, since denominator of the second equation is larger than that of first equation. 

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