Q: A particle is executing SHM with time period T. Starting from mean position, time taken by it to complete 5/8 oscillations, is

Total distance covered by the particle\$ = 4A\$. we divide this whole path in 8 intervals of \$A/2\$. so, 5/8 oscillations means, it has already completed\$1/2\$ oscillation(i.e. total dist. = 2A) and is half way to the other side i.e. A/2. so,\$ A/2 = Asinwt, w= 2pi/ T\$ , substitute to get \$t= T/12.\$ now, total time taken = time to complete previous one half(2A) + time taken to complete\$ A/2 = t/2 + t/12 = 7T/12.\$

Q: The damping force on a oscillator is directly proportional to the velocity. The Unit of the constant of proportionality are:

\$F=Kv\$\$K=\dfrac{F}{V}\$Unit of K = \$K=\dfrac{kg ms^{-2}}{ms^{-1}}=kgs^{-1}\$

Q: The potential energy of a particle of mass \$1\ kg\$ in motion along the \$x\$-axis is given by \$U = 4(1 - \cos 2x)\ J\$. Here \$x\$ is in meter. The period of small oscillations (in sec) is _______.

\$M=1\ kg\$\$U=4(1-\cos2 x)\$\$F=-\dfrac {du}{dx}\$\$F=-4 [0-(-\sin 2x) (2)]\$\$F=-8\sin 2x\ N\$\$a=-8\sin 2x \ m/ \sec^2 \quad [\because \ m=1\ kg]\$\$\Rightarrow \ a\simeq -8(2x)\$\$a=-16x\$ [For small \$x, \sin x \approx x\$]\$\Rightarrow \ \omega^2 =4^2\$\$\omega=4\ \$ rad /ac\$T=\dfrac {2\pi}{w}=\dfrac {\pi}{2}\sec\$

Q: The amplitude of a damped oscillator becomes half on one minute. The amplitude after 3 minute will be \$\displaystyle\dfrac{1}{X}\$ times the original, where \$X\$ is

Amplitude of damped oscillator\$A=A_0e^{- \lambda t} ; \lambda =constant, t=time\$For \$t=1\ min. \dfrac{A_0}{2}=A_0 e^{- \lambda t} \Rightarrow e^{\lambda}=2\$For \$t=3\ min.\ A=A_0e^{- \lambda \times 3 }=\dfrac{A_0}{(e^{\lambda})^3}=\dfrac{A_0}{2^3}\$\$\Rightarrow X=2^3\$

Q: A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to \$\dfrac{1}{1000} \$ of the original amplitude is close to :-

\$A = A_0 e^{-\gamma t}\$\$A = \dfrac{A_0}{2}\$ after 10 oscillations\$\because \$ After 2 seconds\$\dfrac{A_0}{2} = A_0 e^{-\gamma (2)}\$\$2 = e^{2 \gamma}\$\$\ell n 2 = 2 \gamma \$\$\gamma = \dfrac{\ell n 2}{2}\$\$\because A = A_0 e^{-\gamma t}\$\$\ell n \dfrac{A_0}{A} = \gamma t\$\$\ell n 1000 = \dfrac{\ell n 2}{2} t\$\$2 \left(\dfrac{3 \ell n 10}{\ell n2} \right) = t\$\$\dfrac{6 \ell n 10}{\ell n 2} = 1\$\$t = 19.931 sec\$\$t \approx 20 \, sec\$

Q: A particle of mass \$'m'\$ is attached to three identical springs \$A, B\$ and \$C\$ each of force constant \$'k'\$ as shown in figure. If the particle of mass \$'m'\$ is pushed sightly against the spring \$'A'\$ and released. Find the period of oscillation.

\${F_{net}} = {F_A} + {F_B}\cos {45^0} + {F_C}\cos {45^0}\$\$ = ky + 2ky'\cos {45^0} = ky + 2k\left( {y\cos {{45}^0}} \right)\cos {45^0} = 2ky\$\${F_{net}} = k'y \Rightarrow k'y = 2ky \Rightarrow k' = 2k\$\$T = 2\pi \sqrt {\dfrac{m}{{k'}}} = 2\pi \sqrt {\dfrac{m}{{2k}}} \$we get, period of oscillation is \$2\pi \sqrt {\dfrac{m}{{2k}}} .\$

Q: A particle suspended from a fixed point, by a light inextensible thread of length L is projected horizontally from its lowest position with velocity \$\sqrt{\dfrac{7gL}{2}}\$. The thread will slack after swinging through an angle \$\theta\$, such that \$\theta\$ equal.

As shown in figure, \$h=L+\sin\phi\$By energy conservation,\$\dfrac{1}{2}mv^2=mgh+\dfrac{1}{2}mv'^2\$\$\dfrac{1}{2}\times (\sqrt{\dfrac{7gl}{2}})^2=g(L+L\sin\phi)+\dfrac{1}{2}v'^2\$\$v'^2=-2gL-2gL\sin\phi+\dfrac{7}{2} gL\$\$=\dfrac{3}{2}gL-2gL\sin\phi\$If string slack then,\$mg\sin\phi=\dfrac{mv'^2}{L}\$\$g\sin\phi=\dfrac{1}{L}(v')^2=\dfrac{1}{L}(\dfrac{3}{2}gL-2gL\sin\phi)\$\$g\sin\phi=\dfrac{3}{2}g-2g\sin\phi\$\$3g\sin\phi=\dfrac{3}{2}g\$\$\sin\phi=\dfrac{1}{2}\$\$\phi=\dfrac{\pi}{6}\$\$\theta=\dfrac{\pi}{2}+\dfrac{\pi}{6}=\dfrac{4\pi}{6}=120^\circ\$

Q: A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be :

For a spring-mass damped oscillator\$\dfrac{d^2x}{dt^2}+\left(\dfrac{2x}{2\sqrt{mk}}\sqrt{\dfrac{k}{m}}\right)\dfrac{dx}{dt}+\left(\sqrt{\dfrac{k}{m}}\right)^2x=0\$\$\Rightarrow \boxed{\dfrac{d^2x}{dt^2}+2\tau w_0\dfrac{dx}{dt}+w_0^2x=0}\$\$\tau=\dfrac{c}{2\sqrt{mk}}\Rightarrow \$ damping ratio \$c=b\Rightarrow\$ damping constant\$W_0=\sqrt{\dfrac{k}{m}}\Rightarrow\$ angular frequency\$\Rightarrow \boxed{\dfrac{d^2x}{dt^2}+(\dfrac{b}{m})\dfrac{dx}{dt}+(\dfrac{k}{m})x=0}\$..........(1)For LCR circuituse KVL:-\$V-L\dfrac{dI}{dt}-IR-\dfrac{q}{c}=0\$\$\Rightarrow \boxed{\dfrac{d^2q}{dt^2}+\dfrac{R}{L}\dfrac{dq}{dt}+(\dfrac{1}{LC})q=\dfrac{V}{L}}\$...........(2) \$I=\dfrac{dq}{dt}\$By comparing (1) and (2):=\$L\rightarrow m,\ \ \ R\rightarrow b,\ \ \ \ \ LC\rightarrow \dfrac{m}{k}\$ \$\Rightarrow C\rightarrow \dfrac{1}{k}\$option (D) is correct.

Question 1

A particle is executing SHM with time period T. Starting from mean position, time taken by it to complete 5/8 oscillations, is

Accepted Answer

Total distance covered by the&#10;particle$ = 4A$.&#10;&#10;we&#10;divide this whole path in 8 intervals of $A/2$.&#10;&#10;so,&#10;5/8 oscillations means, it has already completed$1/2$ oscillation(i.e. total&#10;dist. = 2A) and is half way to the other side i.e. A/2.&#10;&#10;so,$ A/2 = Asinwt, w= 2pi/ T$ ,&#10;&#10;substitute&#10;to get $t= T/12.$&#10;&#10;now,&#10;total time taken = time to complete previous one half(2A) + time taken to complete$ A/2 = t/2 + t/12 = 7T/12.$

Question 2

The damping force on a oscillator is directly proportional to the velocity. The Unit of the constant of proportionality are:

Accepted Answer

$F=Kv$$K=\dfrac{F}{V}$Unit of K = $K=\dfrac{kg ms^{-2}}{ms^{-1}}=kgs^{-1}$

Question 3

The potential energy of a particle of mass $1\ kg$ in motion along the $x$-axis is given by $U = 4(1 - \cos 2x)\ J$. Here $x$ is in meter. The period of small oscillations (in sec) is _______.

Accepted Answer

$M=1\ kg$$U=4(1-\cos2 x)$$F=-\dfrac {du}{dx}$$F=-4 [0-(-\sin 2x) (2)]$$F=-8\sin 2x\ N$$a=-8\sin 2x \ m/ \sec^2 \quad [\because \ m=1\ kg]$$\Rightarrow \ a\simeq -8(2x)$$a=-16x$ [For small $x, \sin x \approx x$]$\Rightarrow \ \omega^2 =4^2$$\omega=4\ $ rad /ac$T=\dfrac {2\pi}{w}=\dfrac {\pi}{2}\sec$

Question 4

The amplitude of a damped oscillator becomes half on one minute. The amplitude after 3 minute will be $\displaystyle\dfrac{1}{X}$ times the original, where $X$ is

Accepted Answer

Amplitude of damped oscillator$A=A_0e^{- \lambda t} ; \lambda =constant, t=time$For $t=1\ min. \dfrac{A_0}{2}=A_0 e^{- \lambda t} \Rightarrow e^{\lambda}=2$For $t=3\ min.\ A=A_0e^{- \lambda \times 3 }=\dfrac{A_0}{(e^{\lambda})^3}=\dfrac{A_0}{2^3}$$\Rightarrow X=2^3$

Question 5

A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to $\dfrac{1}{1000} $ of the original amplitude is close to :-

Accepted Answer

$A = A_0 e^{-\gamma t}$$A = \dfrac{A_0}{2}$ after 10 oscillations$\because $ After 2 seconds$\dfrac{A_0}{2} = A_0 e^{-\gamma (2)}$$2 = e^{2 \gamma}$$\ell n 2 = 2 \gamma $$\gamma = \dfrac{\ell n 2}{2}$$\because A = A_0 e^{-\gamma t}$$\ell n \dfrac{A_0}{A} = \gamma t$$\ell n 1000 = \dfrac{\ell n 2}{2} t$$2 \left(\dfrac{3 \ell n 10}{\ell n2} \right) = t$$\dfrac{6 \ell n 10}{\ell n 2} = 1$$t = 19.931 sec$$t \approx 20 \, sec$

Question 6

A particle of mass $'m'$ is attached to three identical springs $A, B$ and $C$ each of force constant $'k'$ as shown in figure. If the particle of mass $'m'$ is pushed sightly against the spring $'A'$ and released. Find the period of oscillation.

Accepted Answer

${F_{net}} = {F_A} + {F_B}\cos {45^0} + {F_C}\cos {45^0}$$ = ky + 2ky'\cos {45^0} = ky + 2k\left( {y\cos {{45}^0}} \right)\cos {45^0} = 2ky$${F_{net}} = k'y \Rightarrow k'y = 2ky \Rightarrow k' = 2k$$T = 2\pi \sqrt {\dfrac{m}{{k'}}}  = 2\pi \sqrt {\dfrac{m}{{2k}}} $we get, period of oscillation is $2\pi \sqrt {\dfrac{m}{{2k}}} .$

Question 7

A particle suspended from a fixed point, by a light inextensible thread of length L is projected horizontally from its lowest position with velocity $\sqrt{\dfrac{7gL}{2}}$. The thread will slack after swinging through an angle $\theta$, such that $\theta$ equal.

Accepted Answer

As shown in figure, $h=L+\sin\phi$By energy conservation,$\dfrac{1}{2}mv^2=mgh+\dfrac{1}{2}mv'^2$$\dfrac{1}{2}\times (\sqrt{\dfrac{7gl}{2}})^2=g(L+L\sin\phi)+\dfrac{1}{2}v'^2$$v'^2=-2gL-2gL\sin\phi+\dfrac{7}{2} gL$$=\dfrac{3}{2}gL-2gL\sin\phi$If string slack then,$mg\sin\phi=\dfrac{mv'^2}{L}$$g\sin\phi=\dfrac{1}{L}(v')^2=\dfrac{1}{L}(\dfrac{3}{2}gL-2gL\sin\phi)$$g\sin\phi=\dfrac{3}{2}g-2g\sin\phi$$3g\sin\phi=\dfrac{3}{2}g$$\sin\phi=\dfrac{1}{2}$$\phi=\dfrac{\pi}{6}$$\theta=\dfrac{\pi}{2}+\dfrac{\pi}{6}=\dfrac{4\pi}{6}=120^\circ$

Question 8

Resonance is a special case of forced vibration in which the natural frequency of vibration of the body is the same as the impressed frequency of external periodic force and the amplitude of forced vibration is maximum.

Accepted Answer

The amplitude of forced vibrations of a body increases with an increase in the frequency of the externally impressed periodic force.

Question 9

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be :

Accepted Answer

For a spring-mass damped oscillator$\dfrac{d^2x}{dt^2}+\left(\dfrac{2x}{2\sqrt{mk}}\sqrt{\dfrac{k}{m}}\right)\dfrac{dx}{dt}+\left(\sqrt{\dfrac{k}{m}}\right)^2x=0$$\Rightarrow \boxed{\dfrac{d^2x}{dt^2}+2\tau w_0\dfrac{dx}{dt}+w_0^2x=0}$$\tau=\dfrac{c}{2\sqrt{mk}}\Rightarrow $ damping ratio $c=b\Rightarrow$ damping constant$W_0=\sqrt{\dfrac{k}{m}}\Rightarrow$ angular frequency$\Rightarrow \boxed{\dfrac{d^2x}{dt^2}+(\dfrac{b}{m})\dfrac{dx}{dt}+(\dfrac{k}{m})x=0}$..........(1)For LCR circuituse KVL:-$V-L\dfrac{dI}{dt}-IR-\dfrac{q}{c}=0$$\Rightarrow \boxed{\dfrac{d^2q}{dt^2}+\dfrac{R}{L}\dfrac{dq}{dt}+(\dfrac{1}{LC})q=\dfrac{V}{L}}$...........(2)     $I=\dfrac{dq}{dt}$By comparing (1) and (2):=$L\rightarrow m,\ \ \ R\rightarrow b,\ \ \ \ \ LC\rightarrow \dfrac{m}{k}$                      $\Rightarrow C\rightarrow \dfrac{1}{k}$option (D) is correct.

Question 10

What is meant by Resonance?

Accepted Answer

When the frequency of an externally applied periodic force an a body is equal to the natural frequency of the body, the body readily begins to vibrate or free to vibrate with an increased amplitude. This phenomenon is known as resonance.

Introduction to free and Damped Oscillations

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Examples of Free and Forced Oscillations

Free and Forced Oscillations

Free, Forced, Damped Oscillations and Resonance - Basic

Free and Damped Vibrations

QUICK SUMMARY WITH STORIES

Free and Forced Oscillations

Examples of Free and Forced Oscillations

Damped Vibrations

A particle is executing SHM with time period T. Starting from mean position, time taken by it to complete 5/8 oscillations, is

The damping force on a oscillator is directly proportional to the velocity. The Unit of the constant of proportionality are:

The potential energy of a particle of mass $1 k g$ in motion along the $x$ -axis is given by $U = 4 (1 - cos 2 x) J$ . Here $x$ is in meter. The period of small oscillations (in sec) is _______.

The amplitude of a damped oscillator becomes half on one minute. The amplitude after 3 minute will be $\frac{1}{X}$ times the original, where $X$ is

A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to $\frac{1}{1 0 0 0}$ of the original amplitude is close to :-

A particle of mass $^{'} m^{'}$ is attached to three identical springs $A, B$ and $C$ each of force constant $^{'} k^{'}$ as shown in figure. If the particle of mass $^{'} m^{'}$ is pushed sightly against the spring $^{'} A^{'}$ and released. Find the period of oscillation.

A particle suspended from a fixed point, by a light inextensible thread of length L is projected horizontally from its lowest position with velocity $\frac{7 g L}{2}$ . The thread will slack after swinging through an angle $θ$ , such that $θ$ equal.

Resonance is a special case of forced vibration in which the natural frequency of vibration of the body is the same as the impressed frequency of external periodic force and the amplitude of forced vibration is maximum.

The amplitude of forced vibrations of a body increases with an increase in the frequency of the externally impressed periodic force.

Resonance is a special case of forced vibration in which the natural frequency of vibration of the body is the same as the impressed frequency of external periodic force and the amplitude of forced vibration is maximum.

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be :

What is meant by Resonance?

Introduction to free and Damped Oscillations

Language

Rate

REVISE WITH CONCEPTS

Examples of Free and Forced Oscillations

Free and Forced Oscillations

Free, Forced, Damped Oscillations and Resonance - Basic

Free and Damped Vibrations

QUICK SUMMARY WITH STORIES

Free and Forced Oscillations

Examples of Free and Forced Oscillations

Damped Vibrations

A particle is executing SHM with time period T. Starting from mean position, time taken by it to complete 5/8 oscillations, is

The damping force on a oscillator is directly proportional to the velocity. The Unit of the constant of proportionality are:

The potential energy of a particle of mass 1 kg in motion along the x-axis is given by U=4(1−cos2x) J. Here x is in meter. The period of small oscillations (in sec) is _______.

The amplitude of a damped oscillator becomes half on one minute. The amplitude after 3 minute will be X1​ times the original, where X is

A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 10001​ of the original amplitude is close to :-

A particle of mass ′m′ is attached to three identical springs A,B and C each of force constant ′k′ as shown in figure. If the particle of mass ′m′ is pushed sightly against the spring ′A′ and released. Find the period of oscillation.

A particle suspended from a fixed point, by a light inextensible thread of length L is projected horizontally from its lowest position with velocity 27gL​​. The thread will slack after swinging through an angle θ, such that θ equal.

Resonance is a special case of forced vibration in which the natural frequency of vibration of the body is the same as the impressed frequency of external periodic force and the amplitude of forced vibration is maximum.

The amplitude of forced vibrations of a body increases with an increase in the frequency of the externally impressed periodic force.

Resonance is a special case of forced vibration in which the natural frequency of vibration of the body is the same as the impressed frequency of external periodic force and the amplitude of forced vibration is maximum.

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be :

What is meant by Resonance?

The potential energy of a particle of mass $1 k g$ in motion along the $x$ -axis is given by $U = 4 (1 - cos 2 x) J$ . Here $x$ is in meter. The period of small oscillations (in sec) is _______.

The amplitude of a damped oscillator becomes half on one minute. The amplitude after 3 minute will be $\frac{1}{X}$ times the original, where $X$ is

A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to $\frac{1}{1 0 0 0}$ of the original amplitude is close to :-

A particle of mass $^{'} m^{'}$ is attached to three identical springs $A, B$ and $C$ each of force constant $^{'} k^{'}$ as shown in figure. If the particle of mass $^{'} m^{'}$ is pushed sightly against the spring $^{'} A^{'}$ and released. Find the period of oscillation.

A particle suspended from a fixed point, by a light inextensible thread of length L is projected horizontally from its lowest position with velocity $\frac{7 g L}{2}$ . The thread will slack after swinging through an angle $θ$ , such that $θ$ equal.