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.

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.

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>>Class 11
>>Physics
>>Oscillations
>>Fundamentals of Oscillations and Periodic Motion
>> $Period and Frequency in Osc...$

Period and Frequency in Oscillatory Motion

Q: 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\$

Q: 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 \$.

Q: 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\$

Q: 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.

Q: 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.

Q: 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}\$

Q: 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} \$

Physics

definition

Define frequency of Periodic Motion

Frequency is the number of cycles completed per second. It is the reciprocal of period.
Note: The unit representing cycles per second is hertz.

definition

Define Periodic Motion

Periodic motion, motion repeated in equal intervals of time. Periodic motion is performed, for example, by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.
In each case the interval of time for a repetition, or cycle, of the motion is called a period, while the number of periods per unit time is called the frequency.

definition

Relation between frequency and time period

Relation between frequency and time period:

f = \frac{1}{T}

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