Tag: forced vibration

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Questions Related to forced vibration

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

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:

  1. $\omega _1= \omega _2$

  2. $\omega _1> \omega _2$

  3. $\omega _1 < \omega _2$ when damping is small and $\omega _1> \omega _2$ when damping is large

  4. $\omega _1< \omega _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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$.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

If density (D) acceleration (a) and force (F) are taken as basic quantities,then Time period has dimensions

  1. $\dfrac {1} {6}$ in F

  2. $-\dfrac {1} {6}$ in F

  3. $-\dfrac {2} {3}$ in F

  4. All the above are true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The density has dimension ML-3

so take that as D

the acceleration has dimension LT-2

take it as A

the force has dimension MLT-2

Take it as F

so time period has dimension as D-1/6 A -2/3 F1/6

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

The potential energy of a particle of mass $1\ kg$ in motion along the $x-$axis is given by: $U=4(1-\cos 2x)\ l$, where $x$ is in metres. The period of small oscillations (in sec) is: 

  1. $2\pi$

  2. $\pi$

  3. $\dfrac{\pi}{2}$

  4. $\sqrt {2}\pi$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For small oscillations, U = 4(1 - cos(2x)) approx 4(1 - (1 - (2x)^2/2)) = 4(2x^2) = 8x^2. The force F = -dU/dx = -16x. Comparing to F = -kx, k = 16. Period T = 2*pi * sqrt(m/k) = 2*pi * sqrt(1/16) = 2*pi * 1/4 = pi/2.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

A sphere of radius r is kept on a concave mirror of radius of curvature R. The arrangement is kept on a horizontal table (the surface of concave mirror is frictionless and sliding not rolling). If the sphere is displaced from its equilibrium position and left, then it executes S.H.M. The period of oscillation will be  

  1. $\pi \times { \left( \dfrac { (R-r)1.4 }{ g } \right) } $

  2. $2\pi \times  { \left( \dfrac { R-r }{ g } \right) } $

  3. $\sqrt [ 2\pi ]{ \left( \dfrac { r\quad R }{ g } \right) } $

  4. ${ \left( \dfrac { R }{ g\quad r } \right) } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a sphere of radius r rolling/sliding in a concave mirror of radius R, the effective length of the pendulum is (R-r). The time period for a simple pendulum is T = 2*pi * sqrt(L/g). Substituting L = R-r, we get T = 2*pi * sqrt((R-r)/g).

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5 s .In another 10 s it will decrease to $\alpha $ times its original magnitude where $\alpha $ equals :  

  1. 0.7

  2. 0.81

  3. 0.729

  4. 0.6
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a damped oscillator, amplitude A(t) = A0 * e^(-bt). Given A(5) = 0.9 * A0, so e^(-5b) = 0.9. In another 10s (total 15s), A(15) = A0 * e^(-15b) = A0 * (e^(-5b))^3 = A0 * (0.9)^3 = 0.729 * A0.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

Three infinitely long thin wires, each carrying current I in the same direction, are in the $x-y$ plane of a gravity free space. The central wire is along the y-axis while the other two are along $x=\pm\ d$.
(a) Find the locus of the points for which the magnetic field $B$ is zero.
(b) If the central wire is displaced along the z-direction by a small amount and released, show that it will execute simple harmonic motion. If the linear density of the wires is $\lambda$, find the frequency of oscillation.

  1. $\dfrac{1}{4\pi}\sqrt{\dfrac{\mu _oI^2}{\pi\lambda d^2}}$.

  2. $\dfrac{1}{2\pi}\sqrt{\dfrac{\mu _oI^2}{\pi\lambda d^2}}$.

  3. $\dfrac{2}{2\pi}\sqrt{\dfrac{\mu _oI^2}{\pi\lambda d^2}}$.

  4. $\dfrac{3}{2\pi}\sqrt{\dfrac{\mu _oI^2}{\pi\lambda d^2}}$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a complex problem involving magnetic forces and SHM. The force on the central wire displaced by z is F = - (mu0 * I^2 * z) / (pi * d^2). Using F = ma, where m = lambda * L, the frequency is derived from the restoring force constant.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

A student performs an experiment for determination of $\Bigg \lgroup g = \frac{4\pi^2 l}{T^2} \Bigg \rgroup$, l = 1m, and he commits an error of $\Delta l$ For T he takes the time of n oscillations with the stop watch of least count $\Delta T$ and he commits a human error of 0.1 s. For which of the following data, the measurement of g will be most accurate?

  1. $\Delta L = 0.5, \ \Delta T = 0.1 , \ n = 20$

  2. $\Delta L = 0.5, \ \Delta T = 0.1 , \ n = 50$

  3. $\Delta L = 0.5, \ \Delta T = 0.01 , \ n = 20$

  4. $\Delta L = 0.5, \ \Delta T = 0.05 , \ n = 50$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

It is given that $g$ is determined by $:g = \frac{{4{\pi ^2}l}}{{{T^2}}}$

Taking log and differentiating$,$
$ \Rightarrow \frac{{\Delta g}}{g} = \frac{{\Delta l}}{l} + \frac{{2\Delta T}}{T}$ 
Now$,$ $g$ is most accurate in the case in which error in $g\left( {l.e\,\Delta g} \right)$ is minimum$.$ In case $B,$ number of repetitions to perform the experiment is maximum and ${\Delta g}$  is minimum$.$
Hence,
option $(B)$ is correct answer.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

A particle moves such that its acceleration is given by : $\alpha=-\beta(x-2)$
Here :$\beta$ is a positive constant and x the position from oigin. Time period of oscillations is:

  1. $2\pi+\sqrt\beta$

  2. $2\pi +\sqrt { \cfrac { 1 }{ \beta } } $

  3. $2\pi+\sqrt{\beta+2}$

  4. $2\pi +\sqrt { \cfrac { 1 }{ \beta +2} } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The acceleration is given by a = -beta(x-2). This is the standard form of simple harmonic motion a = -omega^2(x-x0), where omega^2 = beta. The time period T is 2*pi/omega, which simplifies to 2*pi/sqrt(beta).