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 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 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 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 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 3

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 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

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 8

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 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

In forced oscillation of a particle the amplitude is maximum for a frequency $\omega_1$ of force, while the energy is maximum for a frequency $\omega_2$ of the force, then:

Accepted Answer

For the amplitude of oscillation and energy to be maximum, the frequency of force must be equal to the initial frequency and this is only possible in resonance. In resonance state $ \omega_1 = \omega_2$.

Resonance and Condition of Resonance

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Resonance in Waves

Effect of Damping on Resonance

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

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

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.

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 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.

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 :

In forced oscillation of a particle the amplitude is maximum for a frequency $ω_{1}$ of force, while the energy is maximum for a frequency $ω_{2}$ of the force, then:

Resonance and Condition of Resonance

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Rate

REVISE WITH CONCEPTS

Resonance in Waves

Effect of Damping on Resonance

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

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

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.

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 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.

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 :

In forced oscillation of a particle the amplitude is maximum for a frequency ω1​ of force, while the energy is maximum for a frequency ω2​ of the force, then:

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.

In forced oscillation of a particle the amplitude is maximum for a frequency $ω_{1}$ of force, while the energy is maximum for a frequency $ω_{2}$ of the force, then: