Tag: horizontal oscillations of a mass attached to a spring

Questions Related to horizontal oscillations of a mass attached to a spring

A spring balance together with a suspended weight of $2.5$kg is dropped from a height of $30$ metres. The reading on the spring balance, while falling, will show a weight of.

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

  1. $2.5$kg

  2. $1.25$kg

  3. $0$kg

  4. $25$kg

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Correct Option: C
Explanation:

Spring balance reads the net force acting on it by the suspended object. While free falling the object is accelerating with acceleration g m/s-2 . With respect to the spring a pseudo force acts on the object which is equal to mg opposite to the weight of the body. The pseudo force balances the weight of the body and the body does not exert any force on the spring. Thus the spring reading will be zero.

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

Show answer
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 $7s$