Tag: oscillations

Quizzes (16) Questions (206)

Questions Related to oscillations

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

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

  1. K

  2. 2K

  3. 4K

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $2K$

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

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

  1. $\dfrac { \pi } { 5 \sqrt { 2 } } s$
  2. $\dfrac { \pi } { 10 \sqrt { 2 } } s$
  3. $\dfrac { \pi } { 5 } s$
  4. $\dfrac { \pi } { 10 } s$
Reveal answer Fill a bubble to check yourself
D Correct answer
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..

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

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 $6.28\ s$?

  1. $1000\ g$
  2. ${10^5 }\ g$
  3. ${10^7}\ g$
  4. ${10^4}\ g$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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 

  1. $\displaystyle \sqrt{K _{1}/K _{1}}$
  2. $\displaystyle K _{1}/K _{2}$
  3. $\displaystyle K _{2}/K _{1}$
  4. $\displaystyle \sqrt{K _{2}/K _{1}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$V _{1}=V _{2}$
$\sqrt{\dfrac{K _{1}}{m}}A _{1}=\sqrt{\dfrac{K _{2}}{m}}A _{2}$
$\dfrac{A _{1}}{A _{2}}=\sqrt{\dfrac{K _{2}}{K _{1}}}$

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

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 $(1/2) = -0.693)$

  1. $7$s
  2. $9$s
  3. $4$s
  4. $11$s
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $7s$

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

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 $500m$)

  1. $mg$
  2. $ma _{o} \sin\theta$
  3. $\left( m g + m a _ { 0 } \sin \theta \right)$

  4. Zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

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

  1. $40 ms^{-1}$
  2. $2.20 ms^{-1}$
  3. $3.15 ms^{-1}$
  4. $10 ms^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using conservation of momentum, the bob's velocity after impact is v/4. To complete a vertical circle, the bob must have a minimum velocity of sqrt(5gl) at the bottom, which is sqrt(5 * 10 * 2) = 10 m/s. Setting v/4 = 10 gives v = 40 m/s.