Tag: physics

Questions Related to physics

A block of mass m is suspended separately by two different spring have time period $ t _1 and t _2 $ . if same mass is connected to parallel combination of both springs , then its time period is given by

physics simple harmonic motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

  1. $ \dfrac {t _1t _2}{t _1 +t _2} $

  2. $ \dfrac {t _1t _2}{\sqrt {t^2 _1+ t^2 _1} } $

  3. $ \sqrt {\dfrac { t _1t _2}{ t _1 +t _2}} $

  4. $\sqrt {(t _1)^2 + (t _2)^2} $

Show answer
Correct Option: D

Two massless springs of force constants ${ k } _{ 1 }$ and ${ k } _{ 2 }$ are joined end to end. The resultant force constant $k$ of the system is

physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

  1. $k=\dfrac { { k } _{ 1 }+{ k } _{ 2 } }{ { k } _{ 1 }{ k } _{ 2 } } $

  2. $k=\dfrac { { k } _{ 1 }-{ k } _{ 2 } }{ { k } _{ 1 }{ k } _{ 2 } } $

  3. $k=\dfrac { { k } _{ 1 }{ k } _{ 2 } }{ { k } _{ 1 }+{ k } _{ 2 } } $

  4. $k=\dfrac { { k } _{ 1 }{ k } _{ 2 } }{ { k } _{ 1 }-{ k } _{ 2 } } $

Show answer
Correct Option: C
Explanation:

In series, resultant force constant is given as
  $\dfrac { 1 }{ { k } _{ eq } } =\dfrac { 1 }{ { k } _{ 1 } } +\dfrac { 1 }{ { k } _{ 2 } } $
$\Rightarrow { k } _{ eq }=\dfrac { { k } _{ 1 }{ k } _{ 2 } }{ { k } _{ 1 }+{ k } _{ 2 } } $

One end of a long metallic wire of length $L$ area of cross-section $A$ and Young's modulus $Y$ is tied to the ceiling. The other end is tied to a massless spring of force constant $k$. A mass $m$ hangs freely from the free end of the spring. It is slightly pulled down and released. Its time period is given by-

physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

  1. $\displaystyle 2\pi \sqrt{\frac{m}{k}}$

  2. $\displaystyle 2\pi \sqrt{\frac{mYA}{kL}}$

  3. $\displaystyle 2\pi \sqrt{\frac{mk}{YA}}$

  4. $\displaystyle 2\pi \sqrt{\frac{m(kL+YA)}{kYA}}$

Show answer
Correct Option: D
Explanation:
$F = \dfrac{YA\Delta l}{L} = k _2 \Delta l$
we can consider the system as two springs in series hence 
$\dfrac{1}{k _{eq}} = \dfrac{1}{k _1} +\dfrac{1}{k _2}$
$=\dfrac{1}{k} + \dfrac{L}{YA} = \dfrac{YAk +kL}{YAk}$

$ T = 2\pi \sqrt{\dfrac{m}{k _{eq}}} = 2\pi \sqrt{\dfrac{m(YAk + kL)}{YAk}}$

The frequency $f$ of vibrations of a mass $m$ suspended from a spring of spring constant $k$ is given by $f = Cm^xk^y$, where $C$ is a dimensionless constant. The values of $x$ and $y$ are respectively:

physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

  1. $\dfrac{1}{2}, \dfrac{1}{2}$

  2. $-\dfrac{1}{2}, -\dfrac{1}{2}$

  3. $\dfrac{1}{2}, -\dfrac{1}{2}$

  4. $-\dfrac{1}{2}, \dfrac{1}{2}$

Show answer
Correct Option: D
Explanation:
We know $F=-KK\Rightarrow dim\left( K \right) =\left[ { MLT }^{ -2 } \right] \left[ { L }^{ -1 } \right] ={ ML }^{ 0 }{ T }^{ -2 }$
$dim\left( M \right) ={ ML }^{ 0 }{ T }^{ 0 }$    $dim\left( f \right) =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 } \right] $
$f={ Cm }^{ x }{ K }^{ y }\Rightarrow { M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 }={ \left[ { ML }^{ 0 }{ T }^{ 0 } \right]  }^{ k }{ \left[ { ML }^{ 0 }{ T }^{ -2 } \right]  }^{ y }$
$\Rightarrow$  Comparing powers of $M,L$ and $T$ gives,
$x+y=0\quad \quad -2y=-1\quad \Rightarrow \quad y=\dfrac { 1 }{ 2 } $
and $x=-1/2$

Frequency of a block in spring-mass system is $\displaystyle \upsilon $, if it is taken in a lift slowly accelerating upward, then frequency will 

physics simple harmonic motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

  1. decrease

  2. increase

  3. remain constant

  4. none

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Correct Option: C
Explanation:

$\omega=2\pi\sqrt{\dfrac{K}{M}}$

Frequency is independent of gravity
Hence it will remain constant

A uniform spring has certain mass suspended from it and it's period of vertical oscillations is ${t} _{1}$. The spring is now cut in $2$ parts having lengths in ratio $1:2$  and these springs are now connected in series and then in parallel. find out the ratio of the time period of these two ossillation?

physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

  1. $1$

  2. $\sin \theta$

  3. $\sqrt {\dfrac {2}{9}}$

  4. $\sqrt {\dfrac {9}{2}}$

Show answer
Correct Option: C
Explanation:
Let $k$ be initial force constant of spring,${k} _{1}$ and ${k} _{2}$ be the force constant of neew springs
We can derive,
$ kl= constant $
$\Rightarrow \dfrac{{x} _{1}}{{x} _{2}}=1/2$
$\Rightarrow \dfrac{{k} _{1}}{{k} _{2}}=2$     ........$(1)$
$so k _1=3k, k _2=3k/2 $
As initially these lengths were in series:
$\dfrac{1}{k}=\dfrac{1}{{k} _{1}}+\dfrac{1}{{k} _{2}}$
$\Rightarrow \dfrac{1}{k' _1}=\dfrac{1}{3{k} _{1}}+\dfrac{2}{3{k} _{1}}$
$\Rightarrow \dfrac{1}{k' _1}=\dfrac{1}{{k} _{1}}$
$\Rightarrow {k'} _{1}= k\ $
When these two stringd are connected in parallel,
${k _2}^{\prime}={k} _{1}+{k} _{2}$

${k _2}^{\prime}=\dfrac{3k}{2}+3k$

${k _2}^{\prime}=\dfrac{9k}{2}$

Time period is 
$\dfrac{{T _1}^{\prime}}{T' _2}=\sqrt{\dfrac{k' _1}{{k' _2}^{\prime}}}$
$\Rightarrow \dfrac{{T _1}^{\prime}}{T' _2}=\sqrt{\dfrac{2}{9}}$

A $1.5$ kg block at rest on a tabletop is attached to a horizontal spring having a spring constant of $19.6$ N/m. The spring is initially unstretched. A constant $20.0$ N horizontal force is applied to the object causing the spring to stretch.Determine the speed of the block after it has moved $0.30$ m from equilibrium if the surface between the block and the tabletop is frictionless.

physics simple harmonic motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

  1. $2.61\ m/s$

  2. $3.61\ m/s$

  3. $7.61\ m/s$

  4. $8.1\ m/s$

Show answer
Correct Option: B
Explanation:

This system will exhibits S.H.M with angular frequency $\omega =\sqrt { \cfrac { k }{ m }  } =\sqrt { \cfrac { 19.6 }{ 1.5 }  } $

$k$= spring constant 
$m$= mass of the body
The string stretched by the maximum A restoring force at maximum stretch= force acting on body
$\Rightarrow kA=F$
$\Rightarrow A=\cfrac { F }{ k } =\cfrac { 20 }{ 19.6 } $
$F=20N$
$K=19.6\quad N/m$
Now if x is displacement from mean position, the velocity is given by:
$v=\omega =\sqrt { ({ A }^{ 2 }-{ x }^{ 2 }) } $
$v=\omega =\sqrt { \cfrac { 19.6 }{ 1.5 } ({ \cfrac { 20 }{ 19.6 }  }^{ 2 }-{ 0.3 }^{ 2 }) } $
$=3.523m/s$

An infinite number of springs having force constants as K, 2K, 4K, 8K, .......$\displaystyle \infty $ respectively are connected in series; then equivalent spring constant is 

physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

  1. K

  2. 2K

  3. $\displaystyle \frac{K}{2}$

  4. $\displaystyle \infty $

Show answer
Correct Option: C
Explanation:

For the springs connected in series

$\dfrac{1}{K _{eq}}=\dfrac{1}{K}+\dfrac{1}{2K}+\dfrac{1}{4K}+\dfrac{1}{8K}+......$
$\dfrac{1}{K _{eq}}=\dfrac{1}{K}(1+\dfrac{1}{2}+\dfrac{1}{4}+.....)$
$\dfrac{1}{K _{eq}}=\dfrac{1}{K}(\dfrac{1}{1-\dfrac{1}{2}})$
$\dfrac{1}{K _{eq}}=\dfrac{1}{K}(2)$
$K _{eq}=\dfrac{K}{2}$

A body of mass $m$ is suspended from a spring of spring constant $k$. A damping force proportional to the velocity exerts itself on the mass. An appropriate representation of the motion is 

physics simple harmonic motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

  1. $ m \dfrac{d^2x}{dt^2} - c \dfrac{dx}{dt} + kx = 0$

  2. $ m \dfrac{d^2x}{dt^2} + c \dfrac{dx}{dt} - kx = 0$

  3. $ m \dfrac{d^2x}{dt^2} - c \dfrac{dx}{dt} - kx = 0$

  4. $ m \dfrac{d^2x}{dt^2} + c \dfrac{dx}{dt} + kx = 0$

Show answer
Correct Option: D
Explanation:
The forces on the mass $m$ are due to spring and damper.

Suppose the mass moves by a distance $x$ from equilibrium and travels with a velocity $\displaystyle \frac{dx}{dt}$, then

Force due to spring is $-kx$ and force due to damper is $\displaystyle -c\frac{dx}{dt}$

By Newton's Second Law of motion, we have $m\displaystyle \frac{d^2x}{dt^2} = -kx-c\frac{dx}{dt}$

Thus, the equation of motion is $\displaystyle m\frac{d^2x}{dt^2}+c\frac{dx}{dt}+kx=0$

A body of mass $m$ attached to the spring experiences a drag force proportional to its velocity and an external force $F(t) = F _o \cos \omega _ot$. The position of the mass at any point in time can be given by:

physics simple harmonic motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

  1. $x(t) = c _1 \sin (\omega t + \phi) + (\dfrac{F _o}{\omega ^2 - \omega _o ^2}) \cos \omega _o t$

  2. $x(t) = c _1 \cos (\omega t + \phi) + (\dfrac{F _o}{\omega ^2 - \omega _o ^2}) \cos \omega _o t$

  3. $x(t) = c _1 \sin (\omega t) + (\dfrac{F _o}{\omega ^2 - \omega _o ^2}) \cos \omega _o t$

  4. $x(t) = c _1 \cos (\omega t) + (\dfrac{F _o}{\omega ^2 - \omega _o ^2}) \cos \omega _o t$

Show answer
Correct Option: A,B,C,D
Explanation:

Initially, we already know that the displacement at any moment (Instant) of time for spring mass system is:

$x=a\sin { (\omega t+\phi ) } $ or,
$x=a\cos { (\omega t+\phi ) } $
And here an external force is experienced thus, all the four options give position at any time.