Q: Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion constant at any point on its path.

For SHM, acceleration , a = -ω²y F = ma = -mω²y now, work , W = F.dy cos180° { because displacement and acceleration are in opposite direction so, cos180° taken } W = ∫mω²y.dy = mω²y²/2 use standard form of SHM , y = Asin(ωt ± Ф) W = mω²A²/2 sin²(ωt ± Ф) We know, Potential energy is work done stored in system . so, P.E = W = mω²A²/2 sin²(ωt ± Ф) again, Kinetic energy , K.E = 1/2mv² , here v is velocity we know, v = ωAcos(ωt ± Ф) so, K.E = mω²A²/2cos²(ωt ± Ф) Total energy = K.E + P.E = mω²A²/2 cos²(ωt ± Ф) + mω²A²/2 sin²(ωt ± Ф) = mω²A²/2 [ cos²(ωt ± Ф) + sin²(ωt ± Ф) ] = mω²A²/2 [ ∵sin²α + cos²α = 1] = mω²A²/2 = constant

Q: A particle is executing SHM along a straight line. Its velocities at distances \$x_1\$ and \$x_2\$ from the mean position are \$V_1\$ and \$V_2\$, respectively. Its time period is:

Using the expressions for velocity for SHM,\$v_1 = \omega \sqrt{A^2 -x_1^2}\$\$v_2 = \omega \sqrt{A^2 -x_2^2}\$Squaring both,\$v_1 ^2 = \omega ^2 ( A^2 -x_1 ^2)\$ ....(3)\$v_2 ^2 = \omega ^2 ( A^2 -x_2 ^2)\$ ....(4)Subtracting (4) from (3), \$v_1 ^2 - v_2 ^2 = \omega ^2 ( x_2 ^2 - x_1 ^2) \$\$ \Rightarrow 4 \pi ^2 ( x_2 ^2 - x_1 ^2) = T^2 (v_1 ^2 - v_2 ^2 )\$\$ \Rightarrow T = 2\pi \sqrt{\dfrac{ x_2 ^2 - x_1 ^2}{v_1 ^2 - v_2 ^2}}\$ where we have used \$\omega = \dfrac{2\pi}{T}\$

Q: A spring having with a spring constant 1200 N m\$\displaystyle ^{-1}\$ is mounted on a horizontal table as shown in the Figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Spring constant, \$k=1200\,m^{-1}\$Mass, \$m=3kg\$Displacement, \$A=2.0\,cm=0.02m\$(i) Frequency of oscillation v, is given by the relation:\$\displaystyle v=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\$where, T is time period\$\therefore \displaystyle v=\frac{1}{2\times 3.14}\sqrt{\frac{1200}{3}} = 3.18 Hz\$Hence, the frequency of oscillations is 3.18 cycles per second.(ii) Maximum acceleration (a) is given by the relation:\$a=\omega^2\,A\$where,\$\omega =\$ Angular frequency \$=\sqrt{\frac{k}{m}}\$\$A\$ = maximum displacement\$\displaystyle \therefore a=\frac{k}{m}A=\frac{1200\times 0.02}{3}=8\, ms^{-2}\$Hence, the maximum acceleration of the mass is \$8.0\,m/s^2\$(iii) Maximum velocity, \$v_{max}=A\omega\$\$\displaystyle =A\sqrt{\frac{k}{m}}=0.02\times \sqrt{\frac{1200}{3}}=0.4\,m/s\$Hence, the maximum velocity of the mass is 0.4 m/s.

Q: Two masses \$m_1\$ and \$m_2\$ are suspended together by a massless spring of constant K as shown in figure. When the masses are in equilibrium, \$m_1\$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of \$m_2\$.

If only mass \${ m }_{ 2 }\$ is present the extension of spring \$l\$ is given by \${ m }_{ 2 }g=kl\Rightarrow l=\cfrac { { m }_{ 2 }g }{ k } \$, where \$k\$=Spring constantNow with \${ m }_{ 1 }\$ and \${ m }_{ 2 }\$ the extension be \$l'\$ \$({ m }_{ 1 }+{ m }_{ 2 })g=kl'\$\$l'=\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } \$When \${ m }_{ 1 }\$ is removed, \$(l'-1)\$ acts as amplitude of oscillation\$A=(l'-l)=(\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } )g-\cfrac { { m }_{ 2 }g }{ k } \$\$\Rightarrow A=\cfrac { { m }_{ 1 }g }{ k } \$Now, if \${ m }_{ 2 }\$ is removed the angular frequency of oscillation \$\omega =\sqrt { \cfrac { k }{ { m }_{ 2 } } } \$(This is same as the case of \${ m }_{ 2 }\$ attach to spring)\$*\$N.B: When \${ m }_{ 1 }\$ is removed the spring mass system will oscillates with the same angular frequency as the case of spring mass system with only mass \${ m }_{ 2 }\$

Q: A simple pendulum performs simple harmonic motion about \$x = 0\$ with an amplitude \$'a'\$ and time period \$'T'\$. The speed of the pendulum at \$x = \dfrac{a}{2}\$ will be:

As simple pendulum performs simple harmonic motion.\$\therefore velocity, v = \omega \sqrt {a^{2} - x^{2}}\$At, \$x = \dfrac {a}{2}\$\$v = \dfrac {2\pi}{T} \sqrt {a^{2} -\left (\dfrac {a}{2}\right )^{2}} = \dfrac {2\pi}{T} \dfrac {\sqrt {3a^{2}}}{2} = \dfrac {\pi a\sqrt {3}}{T}\$

Q: A particle executes linear simple harmonic motion with an amplitude of \$3\ cm\$. When the particle is at \$2\ cm\$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Velocity of particle executing S.H.M \$v=\omega \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \$ and acceleration, \$a=-{ \omega }^{ 2 }x\$ \$x\$= displacement at any time interval\$a\$ = amplitude\$=3cm\$\$\omega\$= angular frequency, \$=\cfrac { 2\pi }{ T } \$\$T\$=Time periodNow, at \$x=2cm\$ \$v=|a|\$ \$\Rightarrow { \omega }^{ 2 }\times 2=\omega \sqrt { { 3 }^{ 2 }-{ 2 }^{ 2 } } \$\$\Rightarrow { \omega }=\cfrac { \sqrt { 5 } }{ 2 } \$\$\Rightarrow \cfrac { 2\pi }{ T } =\cfrac { \sqrt { 5 } }{ 2 } \$\$\Rightarrow T=\cfrac { 4\pi }{ \sqrt { 5 } } \$

Q: Two identical springs of constant are connected in series and parallel as shown in figure. A mass \$m\$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

We know that, \$n=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\$In series, \$k=\dfrac{k_1k_2}{k_1+k_2}=\dfrac{k^2}{2}\$In parallel, \$k=k_1+k_2=2k\$\$\therefore, n_1=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{2m}}\$and \$n_2 = \dfrac{1}{2\pi}\sqrt{\dfrac{2k}{m}}\$hence, \$n_1:n_2=1:2\$

Q: The kinetic energy and potential energy of a particle executing SHM will be equal when displacement (\$amplitude =a\$) is

\$K.E.=P.E.\\ \Rightarrow \frac { 1 }{ 2 } m{ \omega }^{ 2 }({ a }^{ 2 }-{ x }^{ 2 })=\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ x }^{ 2 }\\ \Rightarrow { a }^{ 2 }-{ x }^{ 2 }={ x }^{ 2 }\\ \Rightarrow 2{ x }^{ 2 }={ a }^{ 2 }\\ \Rightarrow x=a/\sqrt { 2 } \$

Q: A simple pendulum takes \$32\$s to complete \$20\$ oscillations, what is the time period of the pendulum\$?\$

Time taken to complete \$20\$ oscillations \$=32\$sTime taken to complete \$1\$ oscillation \$=32/20\$s \$1.6\$sTime period of a pendulum is time taken by it to complete \$1\$ oscillation.Time period of pendulum is \$1.6\$ seconds.

Question 1

Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion constant at any point on its path.

Accepted Answer

For SHM, &#10;&#10;acceleration , a = -&#969;&#178;y &#10;&#10;F = ma = -m&#969;&#178;y &#10;&#10;now, work , W = F.dy cos180&#176; { because displacement and&#10;acceleration are in opposite direction so, cos180&#176; taken } &#10;&#10;W = &#8747;m&#969;&#178;y.dy = m&#969;&#178;y&#178;/2 &#10;&#10;use standard form of SHM , y = Asin(&#969;t &#177; &#1060;) &#10;&#10;W = m&#969;&#178;A&#178;/2 sin&#178;(&#969;t &#177; &#1060;) &#10;&#10;We know, Potential energy is work done stored in system .&#10;&#10;so, P.E = W = m&#969;&#178;A&#178;/2 sin&#178;(&#969;t &#177; &#1060;) &#10;&#10;again, Kinetic energy , K.E = 1/2mv&#178; , here v is velocity &#10;&#10;we know, v = &#969;Acos(&#969;t &#177; &#1060;) &#10;&#10;so, K.E = m&#969;&#178;A&#178;/2cos&#178;(&#969;t &#177; &#1060;) &#10;&#10;Total energy = K.E + P.E &#10;&#10;= m&#969;&#178;A&#178;/2 cos&#178;(&#969;t &#177; &#1060;) + m&#969;&#178;A&#178;/2&#10;sin&#178;(&#969;t &#177; &#1060;) &#10;&#10;= m&#969;&#178;A&#178;/2 [ cos&#178;(&#969;t &#177; &#1060;) +&#10;sin&#178;(&#969;t &#177; &#1060;) ] = m&#969;&#178;A&#178;/2 [ &#8757;sin&#178;&#945; + cos&#178;&#945; = 1] &#10;&#10;= m&#969;&#178;A&#178;/2 = constant

Question 2

A particle is executing SHM along a straight line. Its velocities at distances $x_1$ and $x_2$ from the mean position are $V_1$ and $V_2$, respectively. Its time period is:

Accepted Answer

&#10;&#10;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;Using the&#10;expressions for velocity for SHM,$v_1 = \omega \sqrt{A^2&#10;-x_1^2}$$v_2 = \omega \sqrt{A^2 -x_2^2}$Squaring&#10;both,$v_1 ^2 = \omega ^2 ( A^2 -x_1 ^2)$ ....(3)$v_2 ^2 =&#10;\omega ^2 ( A^2 -x_2 ^2)$ ....(4)Subtracting (4) from (3), $v_1&#10;^2 - v_2 ^2 = \omega ^2 ( x_2 ^2 - x_1 ^2) $$ \Rightarrow &#10;4 \pi ^2 ( x_2 ^2 - x_1 ^2) = T^2 (v_1 ^2 - v_2 ^2 )$$&#10;\Rightarrow T = 2\pi \sqrt{\dfrac{ x_2 ^2 - x_1 ^2}{v_1 ^2 - v_2&#10;^2}}$ where we have used $\omega = \dfrac{2\pi}{T}$&#10;&#10;

Question 3

A spring having with a spring constant 1200 N m$\displaystyle ^{-1}$ is mounted on a horizontal table as shown in the Figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Accepted Answer

Spring constant, $k=1200\,m^{-1}$Mass, $m=3kg$Displacement, $A=2.0\,cm=0.02m$(i) Frequency of oscillation v, is given by the relation:$\displaystyle v=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$where, T is time period$\therefore \displaystyle v=\frac{1}{2\times 3.14}\sqrt{\frac{1200}{3}} = 3.18 Hz$Hence, the frequency of oscillations is 3.18 cycles per second.(ii) Maximum acceleration (a) is given by the relation:$a=\omega^2\,A$where,$\omega =$ Angular frequency $=\sqrt{\frac{k}{m}}$$A$ = maximum displacement$\displaystyle \therefore a=\frac{k}{m}A=\frac{1200\times 0.02}{3}=8\, ms^{-2}$Hence, the maximum acceleration of the mass is $8.0\,m/s^2$(iii) Maximum velocity, $v_{max}=A\omega$$\displaystyle =A\sqrt{\frac{k}{m}}=0.02\times \sqrt{\frac{1200}{3}}=0.4\,m/s$Hence, the maximum velocity of the mass is 0.4 m/s.

Question 4

Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant K as shown in figure. When the masses are in equilibrium, $m_1$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of $m_2$.

Accepted Answer

If only mass ${ m }_{ 2 }$ is present the extension of spring $l$ is given by ${ m }_{ 2 }g=kl\Rightarrow l=\cfrac { { m }_{ 2 }g }{ k } $, where $k$=Spring constantNow with ${ m }_{ 1 }$ and ${ m }_{ 2 }$ the extension be $l'$ $({ m }_{ 1 }+{ m }_{ 2 })g=kl'$$l'=\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } $When ${ m }_{ 1 }$ is removed, $(l'-1)$ acts as amplitude of oscillation$A=(l'-l)=(\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } )g-\cfrac { { m }_{ 2 }g }{ k } $$\Rightarrow A=\cfrac { { m }_{ 1 }g }{ k } $Now, if ${ m }_{ 2 }$ is removed the angular frequency of oscillation $\omega =\sqrt { \cfrac { k }{ { m }_{ 2 } }  } $(This is same as the case of ${ m }_{ 2 }$ attach to spring)$*$N.B: When ${ m }_{ 1 }$ is removed the spring mass system will oscillates with the same angular frequency as the case of spring mass system with only mass ${ m }_{ 2 }$

Question 5

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $'a'$ and time period $'T'$. The speed of the pendulum at $x = \dfrac{a}{2}$ will be:

Accepted Answer

As simple pendulum performs simple harmonic motion.$\therefore velocity, v = \omega \sqrt {a^{2} - x^{2}}$At, $x = \dfrac {a}{2}$$v = \dfrac {2\pi}{T} \sqrt {a^{2} -\left (\dfrac {a}{2}\right )^{2}} = \dfrac {2\pi}{T} \dfrac {\sqrt {3a^{2}}}{2} = \dfrac {\pi a\sqrt {3}}{T}$

Question 6

A particle executes linear simple harmonic motion with an amplitude of $3\ cm$. When the particle is at $2\ cm$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Accepted Answer

Velocity of particle executing S.H.M $v=\omega \sqrt { { a }^{ 2 }-{ x }^{ 2 } } $ and acceleration, $a=-{ \omega  }^{ 2 }x$ $x$= displacement at any time interval$a$ = amplitude$=3cm$$\omega$= angular frequency, $=\cfrac { 2\pi  }{ T } $$T$=Time periodNow, at $x=2cm$ $v=|a|$ $\Rightarrow { \omega  }^{ 2 }\times 2=\omega \sqrt { { 3 }^{ 2 }-{ 2 }^{ 2 } } $$\Rightarrow { \omega  }=\cfrac { \sqrt { 5 }  }{ 2 } $$\Rightarrow \cfrac { 2\pi  }{ T } =\cfrac { \sqrt { 5 }  }{ 2 } $$\Rightarrow T=\cfrac { 4\pi  }{ \sqrt { 5 }  } $

Question 7

Two identical springs of constant are connected in series and parallel as shown in figure. A mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

Accepted Answer

We know that, $n=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$In series, $k=\dfrac{k_1k_2}{k_1+k_2}=\dfrac{k^2}{2}$In parallel, $k=k_1+k_2=2k$$\therefore, n_1=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{2m}}$and $n_2 = \dfrac{1}{2\pi}\sqrt{\dfrac{2k}{m}}$hence, $n_1:n_2=1:2$

Question 8

The kinetic energy and potential energy of a particle executing SHM will be equal when displacement ($amplitude =a$) is

Accepted Answer

$K.E.=P.E.\ \Rightarrow \frac { 1 }{ 2 } m{ \omega  }^{ 2 }({ a }^{ 2 }-{ x }^{ 2 })=\frac { 1 }{ 2 } m{ \omega  }^{ 2 }{ x }^{ 2 }\ \Rightarrow { a }^{ 2 }-{ x }^{ 2 }={ x }^{ 2 }\ \Rightarrow 2{ x }^{ 2 }={ a }^{ 2 }\ \Rightarrow x=a/\sqrt { 2 } $

Question 9

The displacement time graph of a particle executing S.H.M. is as shown in the figure. The corresponding force-time graph of the particle is:

Accepted Answer

Given displace-time graph indicates a SHM with sinusoidal variation.We know, for a SHM, force $F=-\omega^2y$, where $y$ (displacement) still varies sinusoidally with time.So, $\omega^2$ being positive always, force-time graph will be a negative-sinusoid which is depicted in option (C).

Question 10

A simple pendulum takes $32$s to complete $20$ oscillations, what is the time period of the pendulum$?$

Accepted Answer

Time taken to complete $20$ oscillations $=32$sTime taken to complete $1$ oscillation $=32/20$s $1.6$sTime period of a pendulum is time taken by it to complete $1$ oscillation.Time period of pendulum is $1.6$ seconds.