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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 learning how to measure measurement of small and large distances measurement of distance unit of fundamental quantities

Which of the following system of units is not based on unit of mass, length and time?

  1. CGS

  2. FPS

  3. MKS

  4. SI
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The SI unit system is not based on only mass, length and time alone. 
This unit system is consists of seven fundamental units which can be given as mass, length, time, temperature, luminous intensity, mole and current.
The correct answer is option $(D)$
Multiple choice physics learning how to measure measurement of small and large distances measurement of distance unit of fundamental quantities

Match the Column I with Column II.

Column I Column II
(A) Distance between earth and sun (p) Micron
(B) Interatomic distance in a solid (q) Fermi
(C) Size of a nucleus (r) Light year
(D) Wavelength of infrared laser (s) Angstrom

  1. A - p, B - q, C - r, D - s

  2. A - r, B - s, C - q, D - p

  3. A - q, B - p, C - s, D - r

  4. A - s. B - r, C - p, D - q
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Distance between earth and sun - Light year
Interatomic distance in a solid - Angstrom
Size of a nucleus - Fermi
Wavelength of infrared laser - Micron

Multiple choice physics learning how to measure measurement of small and large distances measurement of distance unit of fundamental quantities

Match the Column I with Column II

Column I(Physical quantity) Column II(Name of unit)
(A) Conductance (p) gray
(B) Magnetic induction (q) lumen
(C) Absorbed dose (r) tesla
(D) Luminous flux (s) siemens

  1. A - s, B - r, C - p, D - q.

  2. A - p, B - q, C - r, D - s.

  3. A - q, B - p, C - s, D - r.

  4. A - r, B - s, C - p, D - q.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

1) Conductance (G) is reciprocal of resistance (R).

$\therefore G=\dfrac{1}{R}$
$\therefore G=\dfrac{I}{V}$      $\left[\therefore \text{since}\,R=\dfrac{V}{I}\right]$
And its Unit is Siemens (S)
2) When an emf [electromotive force] is induced in a circuit due to change in the magnetic flux associated with the circuit is called Magnetic Induction.
$B=\dfrac{F}{q\,v\,\sin\theta}$
Unit is Tesla
3) Absorbed Dose (D) is the amount of energy abdorbed per unit mass of tissue.
Unit is GRAY (Gy)
4) Luminous flux is the measurement of brightness of a light source in terms of energy emitted.
SI unit is Lumen.