Q: 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.

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 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.

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} \$

Question 1

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.

Accepted Answer

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.

Question 2

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

Accepted Answer

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

Question 3

Two identical springs of spring constant K are attached to a block of mass m and to fixed supports as shown in Fig. Shown that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations.

Accepted Answer

Toppr Ask Answer

Question 4

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

Accepted Answer

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 $.

Question 5

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

Accepted Answer

$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.

Question 6

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:

Accepted Answer

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$

Question 7

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

Accepted Answer

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.

Question 8

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?

Accepted Answer

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.

Question 9

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.

Accepted Answer

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                               $T= 2\pi \sqrt{\dfrac{m}{k}} = 2 \pi \sqrt{\dfrac{0.025 \ kg}{2.5 \ N/m}}= \boxed{0.63 \ s}$

Question 10

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:

Accepted Answer

$ 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} $

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