Tag: oscillations due to a spring

Questions Related to oscillations due to a spring

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

Show answer
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.

The natural angular frequency of a particle of mass 'm' attached to an ideal spring of force constant 'K' is

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

  1. $\sqrt{\frac{K}{m}}$

  2. $\sqrt{\frac{m}{K}}$

  3. $\left ( \frac{K}{m} \right )^{2}$

  4. $\left ( \frac{m}{K} \right )^{2}$

Show answer
Correct Option: A
Explanation:
Suppose you displace the particle by a distance $'x'$
The spring now exerts a force,
This provides nccenary force for $SHM$
$\Rightarrow \ F=mwe^2x=k2$ ($w:$ natural angular frequency )
$\Rightarrow \ w=\sqrt {K/m}$

Spring in vehicles are introduces to:

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

  1. Reduce

  2. Reduce impluse

  3. Reduce force

  4. Reduce velocity

Show answer
Correct Option: A

A block of mass m is hanging vertically by spring of spring constant k. If the mass is made to oscillate vertically, its total energy is:

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

  1. maximum at the extreme position

  2. maximum at the mean position

  3. minimum at the mean position

  4. same at all positions

Show answer
Correct Option: D
Explanation:

The block executes SHM. In SHM, the total energy remains constant at all positions.