Tag: forced vibrations

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Questions Related to forced vibrations

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.

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

A particle of mass m is in one dimensional potential field and its potential energy is given by the following equation U(x)=${U _0}\left( {1 - \cos \;aX} \right)\;where\;{U _0}$ and $\alpha $ constants.The period of the particle for small oscillations near the equilibrium will be-

  1. $2\pi \sqrt {\dfrac{{m{\alpha ^2}}}{{m{\alpha ^2}{U _0}}}} $

  2. $2\pi \sqrt {m{\alpha ^2}{U _0}} $

  3. $2\pi \sqrt {\dfrac{m}{{{\alpha ^2}{U _0}}}} $

  4. $2\pi \sqrt {\dfrac{{{\alpha ^2}{U _0}}}{m}} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For small oscillations, U(x) is approximated by a quadratic: U(x) approx (1/2)k*x^2. Here, U(x) = U0(1 - cos(alpha*x)) approx U0(alpha^2 * x^2 / 2). Thus k = U0*alpha^2. T = 2*pi*sqrt(m/k) = 2*pi*sqrt(m / (U0 * alpha^2)).

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

The time taken by a particle performing S.H.M. to pass from point $ A  $ to $  B  $ where its velocities are same is $2$ seconds. After another 2 seconds it returns to $ \mathrm{B}  $ . The time period of oscillation is (in seconds):

  1. $2$

  2. $4$

  3. $6$

  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to the question, a and b points are such that they are

 located at same distances from the equilibrium position. 
Here also it is said that velocity is same 
,i.e; not only the magnitude but also the directions are same.
 So, total time period of oscillation $= 2×$( time taken to go from a to b
 + the next time taken to return at b) $= 2×(2+2)= 8$ sec.

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

A student measures the time period of oscillation of a simple pendulum. He uses the data to estimate the acceleration due to gravity 9g) at that place. If the maximum percentage error in measurement of length pendulum and that in time are $ e _{1} $ and $ e _{2} $ respectively then percentage error estimation of ''g'' is :

  1. $

    e _{1}+2 e _{2}

    $

  2. $

    2 e 1+e 2

    $

  3. $

    e 1+e _{2}

    $

  4. $

    e 1-e _{2}

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a simple pendulum, T = 2*pi*sqrt(l/g), so g = 4*pi^2*l / T^2. The relative error is dg/g = dl/l + 2*dT/T. Thus, the percentage error is e1 + 2*e2.

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

Frequency of oscillation of a body is $6\;Hz$ when force $F _1$ is applied and $8\;Hz$ when $F _2$ is applied. If both forces $F _1\;&\;F _2$ are applied together then, the frequency of oscillation is :

  1. $14\;Hz$

  2. $2\;Hz$

  3. $10\;Hz$

  4. $10\surd{2}\;Hz$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to question,

$F _1=-K _1\,x\;\&\;F _2=-K _2\,x$

So $n _1=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _1}{m}}=6\;Hz;$

$n _2=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _2}{m}}=8\;Hz$

Now $F=F _1+F _2=-(K _1+K _2)x$

Therefore $n=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _1+K _2}{m}}$

$\Rightarrow n=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{4\pi^2\,n^2 _1\,m+4\pi^2\,n^2 _2\,m}{m}}$$=\sqrt{n _1^2+n _2^2}=\sqrt{8^2+6^2}$

$=10\;Hz$