Simple Harmonic Motion Concept Page - 8

Example

Springs in rotational SHM

Example:
A uniform rod of length l mass m is fixed at the centre. A spring of spring constant k is connected  to rod and wall as shown in figure. The rod is displaced by small angle θ$\theta$ and released. Find time period of oscillation.Solution:
Let the length of the rod be l$l$.
Torque acting on the rod about the center of the rod is given by
τ=Iα=−(kx)l2$\tau =I\alpha =-(kx){\displaystyle \frac{l}{2}}$
ml212α=−kl2θ(l2)${\displaystyle \frac{m{l}^2}{12}}\alpha =-k{\displaystyle \frac{l}{2}}\theta ({\displaystyle \frac{l}{2}})$
α=−3kmθ$\alpha =-{\displaystyle \frac{3k}{m}}\theta$
⇒ω=3km−−−√$\Rightarrow \omega =\sqrt{{\displaystyle \frac{3k}{m}}}$
T=2πm3k−−−√$T=2\pi \sqrt{{\displaystyle \frac{m}{3k}}}$

Example

Write the force equation for a mass attached to a spring with an additional constant external force acting on it

Example: The spring shows in figure is unscratched when a man starts pulling on the cord. The mass of the block is M. If the man exerts a constant force F$F$, find the amplitude and the time period of the motion of the block.

Solution:
In equilibrium, let x0$x_0$ is the elongatiun then,
F=kx0$F=k{x}_0$
∴$\therefore$   x0=Fk${\displaystyle x_0=\frac{F}{k}}$
This x0$x_0$ is the amplitude
∴$\therefore$   A=x0=Fk${\displaystyle A=x_0=\frac{F}{k}}$
T=2πMk−−−√${\displaystyle T=2\pi \sqrt{\frac{M}{k}}}$

Example

SHM involving buoyant force

Example:
A small ball of density ρ0${\rho }_0$ is released from rest from the surface of a liquid whose density varies with depth h$h$ as ρ=ρ02(a+βh)$\rho ={\displaystyle \frac{{\rho }_0}{2}}(a+\beta h)$.Mass of the ball is m$m$. Find angular frequency of the ball.Solution:
ρ=ρo2(a+βh)$\rho ={\displaystyle \frac{{\rho }_o}{2}}(a+\beta h)$
V=mρo$V={\displaystyle \frac{m}{{\rho }_o}}$
Downward force, F=mg−mρoρo2(a+βh)g$F=mg-{\displaystyle \frac{m}{{\rho }_o}}{\displaystyle \frac{{\rho }_o}{2}}(a+\beta h)g$
F=mg((1−a2)−βh2)$F=mg((1-{\displaystyle \frac{a}{2}})-{\displaystyle \frac{\beta h}{2}})$
So F$F$ is proportional to displacement from center position which is mg(1−a2)$mg(1-{\displaystyle \frac{a}{2}})$ below surface.
Therefore the motion is SHM
A = mg(1−a2)$mg(1-{\displaystyle \frac{a}{2}})$
k=mgβ2$k={\displaystyle \frac{mg\beta }{2}}$
Also, k=mω2$k=m{\omega }^2$=> ω=gβ2−−−√$\omega =\sqrt{{\displaystyle \frac{g\beta }{2}}}$

Example

Solve problems where component of gravitational force under earth's surface contributes to restoring force

The equation of a particle executing simple harmonic motion is x=(5m)sin[πs−1t+π3]$x=(5m)sin[\pi s^{-1}t+{\displaystyle \frac{\pi }{3}}]$. Write down the amplitude, time period and maximum speed. Also find the velocity at t=1$t=1$ s.

Solution : Comparing with equation x=Asin(ωt+δ)$x=Asin(\omega t+\delta )$, we see that

The  amplitude = 5$5$ m,

And time period =2πω=2ππs−1=2s$={\displaystyle \frac{2\pi }{\omega }}={\displaystyle \frac{2\pi }{\pi s^{-1}}}=2s$

The maximum speed =Aω=5m×πs−1=5πms−1$=A\omega =5m\times \pi s^{-1}=5\pi m{s}^{-1}$

The velocity at time t=dxdt=Aωcos(ωt+δ)$t={\displaystyle \frac{dx}{dt}}=A\omega cos(\omega t+\delta )$.

At t=1$t=1$ s,

v=(5m)(πs−1)cos(π+π3)=−5π2ms−1$v=(5m)(\pi s^{-1})cos(\pi +{\displaystyle \frac{\pi }{3}})={\displaystyle \frac{-5\pi }{2}}m{s}^{-1}$.

Definition

Working of a simple pendulum

A simple pendulum has a heavy point mass (known as bob) suspended from a rigid support by a massless and inextensible string. When the bob from its mean position is pulled to one side and then released, the pendulum is set to motion and the bob moves alternately on either side of its mean position.

Definition

Terms related to Simple Pendulum

These are important terms pertaining to simple pendulum:
1. Oscillation: One complete to and fro motion of the pendulum is called one oscillation
 2. Time period: This is the time taken to complete one oscillation.
3. Frequency of oscillation: It is the number of oscillations made in one second.

Definition

Effective Length of Pendulum

It is the distance of point of oscillation  (i.e. the centre of gravity of the bob) from the point of suspension.

Definition

Define Bob

Bob : A simple pendulum consists of a small metallic ball or a piece of stone suspended from a rigid stand by a thread  called the bob of the pendulum.
Time period of pendulum doesn't depend on the size or mass of the pendulum, it depends on the length of the pendulum and factor of earth's gravity.

Definition

Working of Simple Pendulum

Construction : A simple pendulum consists of a small metallic ball or a piece of stone suspended from a rigid stand by a thread [Fig(a)]. The metallic ball is attached to the pendulum called bob.
Working :Fig(a) shows the pendulum at rest in its mean position. When the bob of the pendulum is released after taking it slightly to one side, it begins to move to and fro [Fig (b)]. The to and fro motion of a simple pendulum is an example of a periodic or an oscillatory motion.
The pendulum is said to have completed one oscillation when its bob,starting from its mean position O, moves to A, to B and back to O.
The pendulum also completes one oscillation when its bob moves from one extreme position A to the other extreme position B and comes back to A.

Definition

Describe a Seconds Pendulum

A seconds pendulum is a pendulum whose period is precisely two seconds; one second for a swing in one direction and one second for the return swing. It has a frequency of 1/2 Hz.

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