The displacement of a particle along the x-axis is given by \\(x=a\sin^2\omega t\\). The motion of the particle corresponds to.

Given, \\(X=a\sin^2\omega t\\)or \\(X=a\left(\displaystyle\frac{1-\cos 2\omega t}{2}\right)\\) \\([\because \cos 2\theta =1-2\sin^2\theta]\\)or \\(X=\displaystyle\frac{a}{2}-\frac{a\cos 2\omega t}{2}\\)Now, \\(V=\displaystyle\frac{dx}{dt}=a\omega \sin 2\omega t\\)and \\(a=dv/dt=2a\omega^2 \cos 2\omega t\\)Here, a is not directly proportional to \\((-x)\\) which is a condition for SHM.

Which of the following equation does not represent a simple harmonic motion:

Standard equation of \\(S.H.M\\) \\( \dfrac {d^2 y}{dt^2}=-\omega^2 y\\), is not satisfied by \\(y=a\tan \omega t\\)

The displacement of a particle is represented by the equation \\(y=sin^3(\omega t)\\). The motion is

Given the equation of displacement of the particle, \\(y={ sin }^{ 3 }\omega t\\)We know \\(sin3\theta =3sin\theta -4{ sin }^{ 3 }\theta \\)Hence, \\(y=\frac { (3sin\omega t-4sin3\omega t) }{ 4 } \\ \Rightarrow 4\frac { dy }{ dt } =3\omega cos\omega t-4\times [3\omega cos3\omega t]\\ \Rightarrow 4\times \frac { { d }^{ 2 }y }{ { dt }^{ 2 } } =-3{ \omega }^{ 2 }sin\omega t+12\omega sin3\omega t\\ \Rightarrow \frac { { d }^{ 2 }y }{ { dt }^{ 2 } } =\frac { -3{ \omega }^{ 2 }sin\omega t+12\omega sin3\omega t }{ 4 } \\ \Rightarrow \frac { { d }^{ 2 }y }{ { dt }^{ 2 } } \\) is not proportional to y. Hence, the motion is not SHM. As the expression is involving sine function, hence it will be periodic. Also \\({ sin }^{ 3 }\omega t={ \left( sin\omega t \right) }^{ 3 }\\ ={ [sin(\omega t+2\pi )] }^{ 3 }\\ ={ [sin(\omega t+2\pi /\omega )] }^{ 3 }\\)Hence, \\(y={ sin }^{ 3 }\omega t\\) represents a periodic motion with period \\(2\pi /\omega \\).

A particle is executing \\(SHM\\) about \\(y = 0\\) along \\(y-\\)axis. Its position at an instant is given by \\(y(m) = 5(sin3\pi t+\sqrt { 3 } cos\ 3\pi t).\\) The amplitude of oscillation is

\\(y(m)=5(\sin { 3\pi t } +\sqrt { 3 } \cos { 3\pi t } )\\ y(m)=5\times 2\left( \cfrac { 1 }{ 2 } \times \sin { 3\pi t } +\cfrac { 1 }{ 2 } \times \sqrt { 3 } \cos { 3\pi t } \right) \\ y(m)=10\left( \cos { \cfrac { \pi }{ 3 } } \times \sin { 3\pi t } +\sin { \cfrac { \pi }{ 3 } } \times \cos { 3\pi t } \right) \\ y(m)=10\sin { \left( 3\pi t+\cfrac { \pi }{ 3 } \right) } \\)So, amplitude of oscillation is 10.

A particle executing SHM has a maximum speed of \\(30 cm s^{-1}\\) and a maximum acceleration of \\(60 cm s^{-2}\\). The period of oscillation is:

Maximum velocity \\(V_{max}=Aw=30cm\text{ }s^{-1}\\)Maximum acceleration \\(a_{max}=A^2w=60cm\text{ }s^{-1}\\)Where \\(A=Amplitude\\w=angular\text{ }frequency\\)\\(\therefore A^2w=60\\)\\(\Longrightarrow A\times 30=60\Longrightarrow A=2cm\\)\\(\therefore w=\cfrac{30}{2}s^{-1}=15sec^{-1}\\)\\(\therefore T=\cfrac{2\pi}{w}=\cfrac{2\pi}{15}sec\\)

A uniform thin ring of radius \\(R\\) and mass \\(m\\) suspended in a vertical plane from a point in its circumference its time period of oscillation is

The time period of the disc is \\(2\pi \sqrt {\left( {3r/2g} \right)} \\)We know that the time period of an object\\(,\\)\\(T = 2\pi \sqrt {\left( {1/mgL} \right)} \\)where\\( ,\\)\\(I =\\) moment of inertia from the suspended point\\(L =\\) distance of its centre from suspended point \\(= r\\)we know that\\(,\\) the moment of inertia of disc about its centre \\(= m{r^2}/2\\)using parallel axis theoram the moment of inertia from a point in its periphery\\(,\\)\\(I = m{r^2} + m{r^2}/2 = 3m{r^2}/2\\)putting the values in the above equation we get\\(,\\)\\(2\pi \sqrt {\left( {3m{r^2}/2mgr} \right)} \\)\\(=2\pi \sqrt {\left( {3r/2g} \right)} \\) therefore\\(,\\) the time period of the disc is \\(2\pi \sqrt {\left( {3r/2g} \right)} \\)Hence,option \\((B)\\) is correct answer.

A block of mass \\(m\\) containing a net positive charges \\(q\\) is placed on a smooth horizontal table which terminates in a ventricles wall as shown in figure \\((29.E2)\\). The distance of the block from the wall is \\(d\\). A horizontal electric field \\(E\\) towards right is switched on. Assuming elastic collisions \\((\\) if any\\()\\) find the time periods of the resulting oscillatory motion. Is it a simple harmonic motion?

Acceleration of block is \\(a=\dfrac{qE}{m}\\)Time taken to reach the wall is equal to time taken from wall to reach the starting position since the collision is elastic.\\(d=\dfrac{1}{2}\dfrac{qE}{m}t^2\\)\\(\implies t=\sqrt{\dfrac{2dm}{qE}}\\)Hence, time taken to reach initial point is \\(T=2t\\)Time of oscillation\\(=T=\sqrt{\dfrac{8dm}{qE}}\\)Since, acceleration is constant ,it is not simple harmonic.

A vertical spring stretches \\(3.9 \ cm\\) when a \\(10-g\\) object is hung from it. The object is replaced with a block of mass \\(25\ g\\) that oscillates up and down in simple harmonic motion. Calculate the period of motion.

The spring constant is found from \\(k= \dfrac{F_s}{x}- \dfrac{mg}{x} = \dfrac{(0.010\ kg)(9.80 \ m/s^2)}{3.9 \times 10^{-2} \ m}= 2.5 \ N/m\\)When the object attached to the spring has mass \\(m= 25 \ g\\), the period of oscillation is \\(T= 2\pi \sqrt{\dfrac{m}{k}} = 2 \pi \sqrt{\dfrac{0.025 \ kg}{2.5 \ N/m}}= \boxed{0.63 \ s}\\)

If the displacement, velocity and acceleration of particle in \\(SHM\\) are \\(1\ cm,1\ cm/sec\\) and \\(1\ cm/sec^{2}\\) respectively its time period (in secs) will be:

\\( x = 1\,cm \\)\\( v = 1\,cm/s \\)\\( a = 1\,cm/s \\)We know that \\( a = -w^{2}x \\)\\( \Rightarrow |a| = w^{2}|x| \\)\\( \Rightarrow w^{2} = \dfrac{|a|}{|x|} = \dfrac{1\,cm/s^{2}}{1\,cm} = 1 \dfrac{1}{s^{2}} \\)\\( \Rightarrow \boxed{w = 1\,rad/s} \\)time period \\( \boxed {T = \dfrac{2\pi }{w} = \dfrac{2\pi }{1} = 2\pi \,sec} \\)

Non-harmonic Oscillations | Definition, Examples, Diagrams

definition

Non-Harmonic Oscillator

When introducing oscillations at the beginning typically only discuss the harmonic oscillator. This is natural, but this focus may disguise its special features (e.g., frequency amplitude independence). Using the air track, our students investigate oscillations experimentally and match experimental data with theoretical. An additional theoretical choice, along with the damped harmonic oscillator, is the solution for the non-harmonic force law:

F = F_{0} s i n (x) v

When there is no damping (=0) there is a difference of less than 0.06 between the non-harmonic and the harmonic oscillator curves if both have unit amplitude and the same frequency and phase. Still, the accuracy of the data allows discrimination between these models. However, the difference shows up most clearly when damping is present since then the period of the non-harmonic oscillator changes with the amplitude and curves that start in phase become out of phase. A simple approximate solution for this damped non harmonic oscillator has been obtained.

Non-harmonic Oscillations

definition

Non-Harmonic Oscillator

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Introduction to Oscillations

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Period and Frequency in Oscillatory Motion

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