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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).

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

Find the time period of small oscillations of the following systems. 

  1. A metre stick suspended through the 20 cm mark.

  2. A ring of mass m and radius r suspended through a point on its perphery.

  3. A uniform square plate of edge a suspended through a corner.

  4. A uniform disc of mass m and radius r suspended through a point r/2 away from the centre.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A physical pendulum's time period is T = 2*pi*sqrt(I/mgd). Option A describes a physical pendulum where the moment of inertia and distance from the center of mass can be calculated to find the period.