Tag: simple harmonic motion

Questions Related to simple harmonic motion

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. ${10^4}\ g$

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

Show answer
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 $7s$

 The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10\ cm$ to $8\ cm$ In $40$ seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$, the time In which amplitude of this pendulum will reduce from $10\ cm$ to $5\ cm$ in carbondioxide will be close to (in $5=1.601, \ln { 2 }  2=0.693$)

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $231\ s$

  2. $208\ s$

  3. $161\ s$

  4. $142\ s$

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Correct Option: C

The time taken for 20 complete oscillations by a seconds pendulum is :

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. 20 s

  2. 50 s

  3. 40 s

  4. 5 s

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Correct Option: C
Explanation:

Time taken for 1 oscillation is 2 s
Time taken for 20 oscillations is $(2 \times 20) = 40  s$

A hollow pendulum bob filled with water has a small hole at the bottom through which water escapes at a constant rate. Which of the following statements describes the variation of the time period (T) of the pendulum as the water flows out?

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. T decreases first and then increases.

  2. T increases first and then decreases.

  3. T increases throughout.

  4. T does not change.

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Correct Option: B
Explanation:

$\displaystyle T = 2\pi \sqrt{\frac{l}{g}}$
First distance of comfrom suspension point will increase then decrease.

The frequency of a second's pendulum is

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. 0.5 Hz

  2. 1.0 Hz

  3. 1.5 Hz

  4. none of these

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Correct Option: A
Explanation:

Seconds pendulum takes 2 $sec$ to make one complete oscillation. 

Hence, frequency = $\dfrac{1}{2}$ Hz

The frequency of a second's pendulum is :

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $0.5\;Hz$

  2. $1.0\;Hz$

  3. $2.0\;Hz$

  4. $5.5\;Hz$

Show answer
Correct Option: A
Explanation:

A seconds pendulum is a pendulum whose period is precisely two seconds