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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}$

$=10\;Hz$

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

The angular frequency of the damped oscillator is given by $\omega =\sqrt { \left( \dfrac { k }{ m } -\dfrac { { r }^{ 2 } }{ 4{ m }^{ 2 } }  \right)  }$ , where k is the spring constant, $m$ is the mass of the oscillator and $r$ is the damping constant. If the ratio $\dfrac { { r }^{ 2 } }{ mk }$ is $80$%, the change in time period compared to the undamped oscillator is approximately as follows:

  1. Decreases by $1$%

  2. Increases by $8$%

  3. Increases by $1$%

  4. Decreases by $8$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The damped frequency is omega' = sqrt(omega0^2 - gamma^2). The time period T' = 2*pi/omega'. For small damping, T' approx T(1 + gamma^2 / (2*omega0^2)). With r^2/mk = 0.8, the change is small and negative/positive depending on the exact definition.

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 - cos2x) J. Here x is in meter. 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
$M=1\ kg$
$U=u(1-\cos^2 x)$
OR $U=8\sin^2 x\quad [\because \ 1-\cos \theta =2\sin^2 \theta]$
$f=-\dfrac {du}{dx}$
$f=-u [0-(-\sin 2x) (2)]$
$f=-8\sin 2x\ N$
$a=-8\sin 2x \ m/ \sec^2 \quad [\because \ m=1\ kg]$
$\Rightarrow \ a\simeq -8(2x)$
$a=-16x$ [For small $x, \sin x \approx x$]
$\Rightarrow \ w^2 =4^2$
$w=4\ $ rad /ac
$T=2\dfrac {\pi}{w}=\dfrac {\pi}{2}\sec$
Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

The period of oscillation of a simple pendulum of constant length is independent of

  1. size of the bob

  2. shape of the bob

  3. mass of bob

  4. all of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$T=2 \pi \sqrt{\dfrac{L}{g}}$               (where L= length of sting)
From above equation, Time period only depend on the length of the string and g.

Option d