Tag: oscillations

Questions Related to oscillations

When a Spring of constant K  is cut into 2 equal parts then new spring constant of both the parts would be:

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. K

  2. 2K

  3. 4K

  4. None of these

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Correct Option: B
Explanation:

$F=KL\ K=\cfrac { F }{ L } \ K\propto \cfrac { 1 }{ L } $

So when it is cut into two equal parts its length decreases to half & simultaneously spring constant increases to $2K$

Two identical particles each of mass $0.5\ kg$ are interconnected by a light spring of stiffness $100\ N/m,$ time period of small oscillation is

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. $\dfrac { \pi } { 5 \sqrt { 2 } } s$

  2. $\dfrac { \pi } { 10 \sqrt { 2 } } s$

  3. $\dfrac { \pi } { 5 } s$

  4. $\dfrac { \pi } { 10 } s$

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Correct Option: D
Explanation:

We know$:$ 

$\mu  = \dfrac{{{m _1}{m _2}}}{{{m _1} + {m _2}}} = \dfrac{m}{2}$
Now$,$ $T = 2\pi \sqrt {\dfrac{\mu }{k}} $
$T = 2\pi \sqrt {\dfrac{{0.5}}{{2 \times 100}}} $
$ = \dfrac{{2\pi }}{{20}}$
$ = \dfrac{\pi }{{10}}s$
Hence,
option $(D)$ is correct answer..

A $100  g$ mass stretches a particular spring by $9.8\ cm,$ when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be $6.28\ s$?

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. $1000\ g$

  2. ${10^5 }\ g$

  3. ${10^7}\ g$

  4. ${10^4}\ g$

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Correct Option: D
Explanation:

$\begin{array}{l} m=0.1\, \, kg,x=9.8\times { 10^{ -2 } }\, \, m,T=6.28\, \, s \ K=\dfrac { { mg } }{ x } \Rightarrow k=10 \ T=2\pi \sqrt { \dfrac { M }{ K }  } \Rightarrow 6.28=2\times 3.14\sqrt { \dfrac { M }{ { 10 } }  }  \ 1=\dfrac { M }{ { 10 } } \Rightarrow M=10\, \, kg={ 10^{ 4 } }g \end{array}$

Two spring-mass systems support equal mass and have spring constants $\displaystyle K _{1}$ and $\displaystyle K _{2}$. If the maximum velocities in two systems are equal then ratio of amplitude of 1st to that of 2nd is 

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. $\displaystyle \sqrt{K _{1}/K _{1}}$

  2. $\displaystyle K _{1}/K _{2}$

  3. $\displaystyle K _{2}/K _{1}$

  4. $\displaystyle \sqrt{K _{2}/K _{1}}$

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Correct Option: D
Explanation:

Maximum velocity $V _{max}=\omega A$

$\omega=\sqrt{\frac{K}{m}}$
$V _{1}=\sqrt{\dfrac{K _{1}}{m}}A _{1}$
$V _{2}=\sqrt{\dfrac{K _{2}}{m}}A _{2}$
It is given that both have same maximum velocity and same mass
$V _{1}=V _{2}$
$\sqrt{\dfrac{K _{1}}{m}}A _{1}=\sqrt{\dfrac{K _{2}}{m}}A _{2}$
$\dfrac{A _{1}}{A _{2}}=\sqrt{\dfrac{K _{2}}{K _{1}}}$

In the above question, the velocity of the rear 2 kg block after it separates from the spring will be :

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. 0 m/s

  2. 5 m/s

  3. 10 m/s

  4. 7.5 m/s

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Correct Option: A

A block of mass $200$ g executing SHM under the influence of a spring of spring constant $k = 90 N m^{-1}$ and a damping constant $b = 40 g s^{-1}$. Time taken for its amplitude of vibrations to drop to half of its initial values (Given, In $(1/2) = -0.693)$

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. $7$s

  2. $9$s

  3. $4$s

  4. $11$s

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Correct Option: A
Explanation:

Given data,

mass $m=200g$
Spring constant $k=90Nm^{-1}$
Damping constant $b=40gs^{-1}$
To calculate: Time taken for the amplitude of vibration to drop to half of the initial value
We know that amplitude at any time t can be given as:

 $A(t)=A _0e^{-\dfrac{bt}{2m}}$

or $T _{1/2}=\dfrac{-0.693l×2×0.2}{40×10^{−3}}=6.93s$

Time taken for its amplitude of vibrations to drop to half of its initial values is $7s$

A large box is accelerated up the inclined plane with an acceleration a and pendulum is kept vertical (Somehow by an external agent) as shown in figure.Now if the pendulum is set free to oscillate from such position, then what is the tension in the string immediately after the pendulum is set free? (mass of $500m$)

angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

  1. $mg$

  2. $ma _{o} \sin\theta$

  3. $\left( m g + m a _ { 0 } \sin \theta \right)$

  4. Zero

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Correct Option: C

 The time period of oscillation of a torsional pendulum of moment of inertia I is

angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

  1. $T =2 \pi \sqrt{I/k}$

  2. $T =2 \pi \sqrt{I/2k}$

  3. $T =2 \pi \sqrt{2I/k}$

  4. $T =2 \pi \sqrt{I/4k}$

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Correct Option: A
Explanation:

The time period of oscillations of a torsional pendulum is $T =2 \pi \sqrt(I/k)$

The correct option is (a)

A bullet of mass $'m'$ hits a pendulum bob of mass $'2m'$ with a velocity $'v'$ and comes out of the bob with velocity $v/2$. Length of the pendulum is $2$ meter and $g=10 ms^{-2}$. The minimum value of $'v'$ for the bullet so that the bob may complete one revolution in the verticle is

angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

  1. $40 ms^{-1}$

  2. $2.20 ms^{-1}$

  3. $3.15 ms^{-1}$

  4. $10 ms^{-1}$

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Correct Option: A

Time period of a disc about a tangent parallel to the diameter is same as the time period of a simple pendulum. The ratio of radius of disc to the length of pendulum is :

angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

  1. $\dfrac { 1 } { 4 }$

  2. $\dfrac { 4 } { 5 }$

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

  4. $\dfrac { 1 } { 2 }$

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Correct Option: C