Tag: simple harmonic motion

Quizzes (5) Questions (96)

Questions Related to simple harmonic motion

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

 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 $5=1.601, \ln { 2 }  2=0.693$)

  1. $231\ s$
  2. $208\ s$
  3. $161\ s$
  4. $142\ s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The amplitude decay of a pendulum in a viscous medium follows the equation A = A_0 * exp(-bt/2m). The damping constant b is proportional to the viscosity eta. Since the ratio of viscosities is 1.3, the decay constant in CO2 is 1.3 times that in air. By comparing the time taken to reach half amplitude, the result is calculated as 161 seconds.

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

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?

  1. T decreases first and then increases.

  2. T increases first and then decreases.

  3. T increases throughout.

  4. T does not change.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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