Which of the following functions of time represent \\((A)\\) simple harmonic, \\((B)\\) periodic but not simple harmonic, and \\((C)\\) non-periodic motion? Give period for each case of periodic motion ( \\(\displaystyle \omega \\) is any positive constant):(a) \\(\displaystyle \sin \omega t-\cos \omega t\\)(b) \\(\displaystyle \sin ^{3}\omega t\\)(c) \\(\displaystyle 3 \cos \left ( \pi /4-2\omega t \right )\\)(d) \\(\displaystyle \cos \omega t+\cos 3 \omega t +\cos 5 \omega t\\)(e) \\(exp \displaystyle \left ( -\omega ^{2}t^{2} \right )\\)(f) \\(1+\displaystyle \omega t+\omega ^{2}t^{2}\\)

(a) SHMThe given function is:\\(sin\, \omega t - cos\, \omega t\\)\\(=\sqrt{2}\left[\frac{1}{\sqrt{2}}sin\, \omega t - \frac{1}{\sqrt{2}}cos\, \omega t\right]\\)\\(=\sqrt{2}\left[sin\, \omega t\times cos\frac{\pi}{4}-cos\,\omega t\times sin \frac{\pi}{4}\right]\\)\\(=\sqrt{2}\,sin(\omega t-\frac{\pi}{4})\\)This function represents SHM as it can be written in the form: a \\(sin (\omega t+\phi)\\) Its period is: \\(2\pi/\omega\\).(b) Periodic but not SHMThe given function is:\\(sin^3\omega t = 1/4[3\,sin\,\omega t - sin\, 3\omega t]\\)The terms sin ωt and sin 3ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.Its period is: \\(2\pi/\omega\\) (LCM of time periods).(c) SHMThe given function is:\\(3\,cos\left[\frac{\pi}{4} - 2\,\omega t\right]\\)\\(=-3\,cos\left[2\,\omega t - \frac{\pi}{4}\right]\\)This function represents simple harmonic motion because it can be written in the form: a \\(cos(\omega t+\phi)\\). Its period is: \\(2\pi/2\omega =\pi/\omega\\)(d) Periodic, but not SHMThe given function is \\(cos\, \omega t+cos\, 3\, \omega t+cos\,5\omega t\\). Each individual cosine function represents SHM. However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.Its period is LCM of the period of the three sinusoids = \\(2\pi/\omega\\)(e) Non-periodic motionThe given function exp\\((-\omega^2t^2)\\) is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.(f) The given function \\(1+\omega t+\omega^2t^2\\) is non-periodic.

A particle executes a simple harmonic motion of time period \\(T\\). Find the time taken by the particle to go directly from its mean position to half the amplitude.

The displacement equation \\(x = A \\) \\(sin wt\\) where \\(A\\) is the amplitude of oscillation The particle is at mean position initially i.e at \\(t= 0\\), \\(x =0\\)At the instant \\(t\\), the particle is at a position of half amplitude i.e \\(x = \dfrac{A}{2}\\)\\(\therefore\\) \\(\dfrac{A}{2} = A \\) \\(sinwt\\) \\(0.5 = sinwt\\) \\(\implies wt = \dfrac{\pi}{6}\\)\\(\dfrac{2\pi}{T} \times t = \dfrac{\pi}{6}\\) \\(\implies t = \dfrac{T}{12}\\)

The displacement of a particle executing simple harmonic motion is given by \\(y = A_0 + A \sin \omega t + B \cos \omega t\\)Then the amplitude of its oscillation is given by :

Given,\\(y = A_0 + A sin \omega t + B sin \omega t\\)Equation of SHM\\(y' = y - A_0 = A sin \omega t + B cos \omega t\\)Resultant amplitude,\\(R = \sqrt{A^2 + B^2 + 2 AB cos 90^{\circ}}\\)\\(= \sqrt{A^2 + B^2}\\)

Two particle are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is:

HINT: DISTANCE IS DEFINED AS THE AREA COVERED BY AN OBJECT FROM MOVING FROM ONE POINT TO ANOTHER STEP1:In fig (1) particles at a mean position.In fig(2) particles staring some displacement but crossing each other in opposite direction. STEP2:Now if the displacement of two particle \\(\mathrm{x}_{1}=\mathrm{A} \sin (\omega \mathrm{t})\\) \\(\mathrm{x}_{2}=\mathrm{A} \sin (\omega \mathrm{t})\\) For \\(\mathrm{x}_{1}=\dfrac{\mathrm{A}}{2}, \omega \mathrm{t}=\dfrac{\pi}{6} \ldots, \dfrac{5 \pi}{6} \ldots, \ldots\\) \\(\mathrm{x}_{2}=\dfrac{\mathrm{A}}{2}, \omega \mathrm{t}=\dfrac{\pi}{6} ., \dfrac{5 \pi}{6}\\) From fig (2) we can say point \\(A\\) and B corresponding \\(\omega \mathrm{t}=\dfrac{5 \pi}{6}, \omega \mathrm{t}=\dfrac{\pi}{6}\\) So phase difference \\(\left(\dfrac{5 \pi}{6}-\dfrac{\pi}{6}\right)=\dfrac{2 \pi}{3}\\) [N.B \\(\rightarrow\\) Before starting motion of ' 2 '...particle ' 1 ' goes to right extreme end and while '1 returning to mean position particle '2' starts its motion and the point of equal displacements are A and B]

A particle executing SHM of amplitude \\(4 cm\\) and \\(T=4s\\). The time taken by it to move from positive extreme position to half the amplitude is

R.E.F image \\( W = \dfrac{2\pi }{T} = \dfrac{2\pi }{4} = \dfrac{\pi}{2} \\) Using Phalov diagram \\( wt = \dfrac{\pi}{3} \\)\\( \dfrac{\pi}{2}(t) = \dfrac{\pi}{3} \\)\\( \Rightarrow \boxed {t = \dfrac{2}{3}s} \\) Ans

A particle executes simple harmonic oscillation with an amplitude \\(a\\). The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Equation of a SHM is given by \\(X=Asin(\omega t)\\)Here \\(\omega =\frac{2\pi}{T}\\)Time to travel half of the amplitude\\(\dfrac{A}{2}=A sin(\omega t)\\)\\(\Rightarrow \omega t = \dfrac{\pi}{6}\\)\\(\Rightarrow t = \dfrac{T}{12}\\)

Two particles are executing simple harmonic motion of the same amplitude \\(A\\) and frequency \\(\omega\\) along the x-axis. Their mean position is separated by distance \\({X}_{0}({X}_{0}> A)\\). If the maximum separation between them is \\(({X}_{0}+A)\\), the phase difference between their motion is

\\(\begin{array}{l} { x_{ 1 } }=A\sin \left( { \omega t+{ \phi _{ 1 } } } \right) \\ { x_{ 2 } }=A\sin \left( { \omega t+{ \phi _{ 2 } } } \right) \\ { x_{ 1 } }-{ x_{ 2 } }=A\left[ { 2\sin \left[ { \omega t+\dfrac { { { \phi _{ 1 } }+{ \phi _{ 2 } } } }{ 2 } } \right] \sin \left[ { \dfrac { { { \phi _{ 1 } }-{ \phi _{ 2 } } } }{ 2 } } \right] } \right] \\ A=2A\sin \left( { \dfrac { { { \phi _{ 1 } }-{ \phi _{ 2 } } } }{ 2 } } \right) \\ \dfrac { { { \phi _{ 1 } }-{ \phi _{ 2 } } } }{ 2 } =\dfrac { \pi }{ 6 } \\ { \phi _{ 1 } }=\dfrac { \pi }{ 3 } \end{array}\\)Hence,option \\((B)\\) is correct answer.

The time taken by a particle performing SHM to pass from point A to B where its velocities are same is \\(2s\\). After another \\(2s\\), it returns to B.The time period of oscillation is

We see that the velocity at A and B are same and it takes \\(2\\) s to cover this distance.Now it takes \\(2\\) s more to come back to B after travelling to the extreme position.We see that the distance covered from A to B and then from B and then back to B from extreme takes \\(4\\) s. This is half the distance in the whole top and fro path. Thus the time taken for one oscillation should be \\(8\\) s

For a particle executing simple harmonic motion, the displacement x is given by \\(x = A cos \omega t\\). Identify the graph, which represents the variation of potential energy (U) as a function of time t and displacement x.

Given : \\(x = A cos \omega t\\)Potential energy, \\(U = \frac{1}{2} m \omega^2x^2 = \frac{1}{2}m \omega^2 A^2cos^2 \omega t\\)On plotting a graph between potential energy U and time t, we shall get a curve II and graph between U and x we shall get curve IV.

The displacement - time graph for a particle executing SHM is as shown in figureWhich of the following statement is correct?

For the SHM,The displacement of the particle is given by, \\(x = A cos \omega t\\)Velocity, \\(v = \dfrac{dx}{dt} = \dfrac{d}{dt} (A cos \omega t) = -A \omega sin \omega t\\)Acceleration, \\(a = \dfrac{dv}{dt} = \dfrac{d}{dt}(- \omega A sin \omega t) = - \omega^2A cos \omega t\\).The corresponding velocity-time and acceleration time graph is as shown in the figure.Option \\(A\\)

Displacement function in SHM | Formulas, Definition, Law

example

Extreme Position(Amplitude) of a particle performing SHM

Example : A particle moves along y-axis according to the equation y (in cm)

= 3 sin 100 π t + 8 sin^{2} 50 π t - 6

.Find whether the motion is simple harmonic or not.Also, calculate amplitude of particle and its mean position.
Solution:
The given equation can be written as

y = 3 sin 100 π t + (4 - 4 cos 100 π t) - 6

= 3 sin 100 π t + 4 sin (100 π t + π / 2) - 2

y = 5 sin (100 π - 5 3^{\circ}) - 2

y_{m a x} = 5 - 2 = 3

y_{m i n} = - 5 - 2 = - 7

Mean position

= \frac{y _{m a x} + y _{m i n}}{2} = - 2

example

Use phase to understand relative position between two SHMs

Example: What is the minimum phase difference between two SHMs

y_{1} = s i n (π / 6) s i n (ω t) + s i n (π / 3) c o s (ω t)

;

y_{2} = c o s (π / 6) s i n (ω t) + c o s (π / 3) c o s (ω t)

?

Solution:

y_{1} = s i n (ω t) s i n (\frac{π}{6}) + c o s

ω t s i n \frac{π}{3}

= s i n

ω t c o s (\frac{π}{3}) + c o s

ω t s i n \frac{π}{3}

= s i n (ω t + \frac{π}{3})

y_{2} = s i n (ω t + \frac{π}{6})

Thus phase difference is

\frac{π}{3} - \frac{π}{6} = \frac{π}{6}

shortcut

Find time taken to travel between two given positions in SHM

Example: A particle executes SHM along a straight line with mean position at

x = 0

and with a period of

20

sec and amplitude of

5

cm. Find the shortest time taken by it to go from

x = 4

cm to

x = - 3

cm ?

Solution:

x = A

sin(

ω t + ϕ

)

ω = \frac{2 π}{2 0} = \frac{2 π}{2 0} = \frac{π}{1 0} \frac{r a d}{s e c}

let at

t = 0, x = 4

, thus

4 = 5

sin

ϕ

s i n^{- 1} \frac{4}{5} = ϕ

Now for at

t = t_{1}

, let

x = - 3

, thus we have

\frac{- 3}{5} = s i n (ω t_{1} + ϕ)

s i n^{- 1} (\frac{- 3}{5}) - s i n^{- 1} (\frac{4}{5}) = ω t_{1}

10 \frac{( - 0 . 6 4 3 - 0 . 9 2 7 )}{π} = t_{1}

Solving we get

t_{1} = 5 s e c

law

Displacement as a function of time is a simple harmonic motion

Standard equation of simple harmonic motion is:

a = - w^{2} x

Any general equation satisfying the above criterion represents a simple harmonic motion.
i.e.

x = A s i n w t

formula

Angular displacement as a function of time

In angular SHM equation of motion is given by:

τ = - k θ

General equation for angular displacement:

θ = θ_{o} s i n (w t + ϕ)

diagram

Shift of displacement-time plot with change in phase

In the given plot, phase difference is

π / 4

x_{1} (t) = A sin (ω t)

x_{2} (t) = A sin (ω t + π / 4)

Displacement function in SHM

example

Extreme Position(Amplitude) of a particle performing SHM

example

Use phase to understand relative position between two SHMs

shortcut

Find time taken to travel between two given positions in SHM

law

Displacement as a function of time is a simple harmonic motion

formula

Angular displacement as a function of time

diagram

Shift of displacement-time plot with change in phase

Quick Summary With Stories

Displacement in SHM

A particle executes a simple harmonic motion of time period $T$ . Find the time taken by the particle to go directly from its mean position to half the amplitude.

The displacement of a particle executing simple harmonic motion is given by $y = A_{0} + A sin ω t + B cos ω t$
Then the amplitude of its oscillation is given by :

A particle executing SHM of amplitude $4 c m$ and $T = 4 s$ . The time taken by it to move from positive extreme position to half the amplitude is

A particle executes simple harmonic oscillation with an amplitude $a$ . The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $ω$ along the x-axis. Their mean position is separated by distance $X_{0} (X_{0} > A)$ . If the maximum separation between them is $(X_{0} + A)$ , the phase difference between their motion is

The time taken by a particle performing SHM to pass from point A to B where its velocities are same is $2 s$ . After another $2 s$ , it returns to B.The time period of oscillation is

For a particle executing simple harmonic motion, the displacement x is given by $x = A c o s ω t$ . Identify the graph, which represents the variation of potential energy (U) as a function of time t and displacement x.

The displacement - time graph for a particle executing SHM is as shown in figure
Which of the following statement is correct?

example

Extreme Position(Amplitude) of a particle performing SHM

example

Use phase to understand relative position between two SHMs

shortcut

Find time taken to travel between two given positions in SHM

law

Displacement as a function of time is a simple harmonic motion

formula

Angular displacement as a function of time

diagram

Shift of displacement-time plot with change in phase

Quick Summary With Stories

Displacement in SHM

A particle executes a simple harmonic motion of time period T. Find the time taken by the particle to go directly from its mean position to half the amplitude.

The displacement of a particle executing simple harmonic motion is given by y=A0​+Asinωt+Bcosωt Then the amplitude of its oscillation is given by :

A particle executing SHM of amplitude 4cm and T=4s. The time taken by it to move from positive extreme position to half the amplitude is

A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x-axis. Their mean position is separated by distance X0​(X0​>A). If the maximum separation between them is (X0​+A), the phase difference between their motion is

The time taken by a particle performing SHM to pass from point A to B where its velocities are same is 2s. After another 2s, it returns to B.The time period of oscillation is

For a particle executing simple harmonic motion, the displacement x is given by x=Acosωt. Identify the graph, which represents the variation of potential energy (U) as a function of time t and displacement x.

The displacement - time graph for a particle executing SHM is as shown in figure Which of the following statement is correct?

A particle executes a simple harmonic motion of time period $T$ . Find the time taken by the particle to go directly from its mean position to half the amplitude.

The displacement of a particle executing simple harmonic motion is given by $y = A_{0} + A sin ω t + B cos ω t$
Then the amplitude of its oscillation is given by :

A particle executing SHM of amplitude $4 c m$ and $T = 4 s$ . The time taken by it to move from positive extreme position to half the amplitude is

A particle executes simple harmonic oscillation with an amplitude $a$ . The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $ω$ along the x-axis. Their mean position is separated by distance $X_{0} (X_{0} > A)$ . If the maximum separation between them is $(X_{0} + A)$ , the phase difference between their motion is

The time taken by a particle performing SHM to pass from point A to B where its velocities are same is $2 s$ . After another $2 s$ , it returns to B.The time period of oscillation is

For a particle executing simple harmonic motion, the displacement x is given by $x = A c o s ω t$ . Identify the graph, which represents the variation of potential energy (U) as a function of time t and displacement x.

The displacement - time graph for a particle executing SHM is as shown in figure
Which of the following statement is correct?