The motion in which the position of a body repeats after fixed interval of time is known as periodic or harmonic motion. A harmonic motion is simplest type i.e. constant amplitude and simple frequency is known as simple harmonic motion. A body moves to and fro about it’s mean position, the acceleration so produced is directly proportional to the displacement 9y0 and is always directed towards the mean position.
$$\text {i.e.} a = -ky $$
Where negative sign shows that the acceleration and displacement are in opposite directions.
Let us consider a particle moving around a circle of radius r with a uniform angular velocity () in anticlockwise direction as shown in the figure. XOX’ and YOY’ are two mutually perpendicular diameters of the circle. As the particle goes in the circle, the foot of the perpendicular along Y-axis executes oscillatory motion.
Let at any time t, the position of the particle is P and angular displacement Ï´. Let M and N be the foot of the perpendicular drawn from P on XOX’ and YOY’ respectively. The displacement of the foot of perpendicular N on diameter YOY’ is ON.
\begin{align*} \sin \theta &= \frac {ON}{OP} \\ As \:ON &= y, OP = r \\ \therefore \sin \theta &= \frac yr \\ \text {or,} \: y &= r\sin \theta \\ \text {or,} \: y &= r\sin \omega t \dots (i) \end{align*}
Where \(\theta = \omega t\). This is the displacement equation for s S.H.M which is periodic, sinusoidal function of time. It can be expressed in terms of cosine function and similar expression can be obtained in any diameter of the circle.
Velocity
It is the rate of change of displacement of a body.
\begin{align*} v &= \frac {dy}{dt} = \frac {d(r\sin \omega t)}{dt} \\ &= r \omega \cos \omega t \dots (ii) \\ &= r\omega \sqrt {1 - \sin ^2 \omega t} \\ &= r\omega \sqrt { 1 - \left (\frac yr \right ) ^2} \\ &= \omega \sqrt {r^2 –y^2} \dots (iii) \end{align*}
$$ \text {Case I, if}\: y = 0 , V_{max} = \omega r$$
$$\text {Case II, If} \: y = r, V_{min} = 0$$
Acceleration
It is the rate of change of velocity of a body.
\begin{align*} a &= \frac {dv}{dt} = r\omega \frac {d}{dt} \cos \omega t \\ &= r\omega \times -\omega \sin \omega t \\ &= \omega ^2 r \sin \omega t \dots (iv) \\ &= -\omega ^2y\dots (v) \\ \end{align*}
Here \(a \propto y\) and is directed toward mean position so the motion is S.H.M.
Amplitude
It is the maximum displacement of a body from its mean position in periodic motion. The displacement in S.H.M. at any time is given by the relation, \( y = r \sin \omega t\). When \(\sin \omega t = 1\), then y_{max} = r. So, the amplitude of motion is r.
Time Period (T)
It is the time required by a body to complete one revolution.
$$ T = \frac {2\pi }{\omega } = 2\pi \sqrt {\frac ya} (\therefore a = \omega ^2y, \omega = \sqrt {\frac ay} )$$
Frequency (F)
It is the number of complete rotations made by a body in 1 second.
$$ F = \frac 1T = \frac {\omega}{2\pi} = 2\pi \sqrt {\frac ay} $$
Phase (Θ)
The phase of the body at any time is defined as the position and direction of its motion with respect to mean position at that time.
Bob of simple pendulum started oscillating simple harmonically from extreme position. After what minimum time ti will be at x = A/2 ?
T/4
T/6
T/2
T/12
If the length of simple pendulum increases by 2% then percentage increase in its time period depends is ______.
A hole is made through centre of earth and a stone is put into the hole. The morion of stone is ______.
linear
none of the answers are correct
simple harmonic motion
stone stops at centre of earth
Bob of simple pendulum is fitted with liquid. If pluged hole at bottom of bob is suddendly unplugged then the time period of simple pendulum till water comes out ______.
decreases
first decreases and increases
first increases then decreases
increases
If frequency of oscillation of particle executing simple harmonic motion in n then the frequency of oscillation of kinetic energy is ______.
n^{2}
2n
n
n/2
Second's Pendulum id defined as
Two simple pendula of lengths 5m and 20 cm, respectively are given small displacement in one direction at the same time from similar position. After how manyminimum oscillation of first pendulum, they will be in phase again?
4
3
2
1
Fraction of kinetic energy of particle executing simple harmonic motion at x=0 (x gives the position of particle) is _______.
2
3/2
1/2
1
The string the spring and the pulley shown in figure are light. Find the time period of the mass m ______.
$$ 4pi sqrt {frac mk} $$
$$ 2pi sqrt {frac km} $$
$$ pi sqrt {frac mk} $$
$$ 2pi sqrt {frac mk} $$
Two simple pendula of lengths 5m and 20 cm, respectively are given small displacement in one direction at the same time from similar position. After how manyminimum oscillation of first pendulum, they will be in phase again?
1
2
3
4
What is time period of oscillation of liquid in U-shaped tube as shown in the figure?
$$2pi ext {sec} $$
$$pi ext {sec} $$
$$frac {pi}{2} ext {sec} $$
$$pi ext {sec} $$
Find the time period of small back and forth on a smooth concave surface of radius R.
$$pi sqrt {frac Rg} $$
$$pi sqrt {frac gR} $$
$$2pi sqrt {frac gR} $$
$$2pi sqrt {frac Rg} $$
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