Tag: horizontal oscillations of a mass attached to a spring

Questions Related to horizontal oscillations of a mass attached to a spring

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

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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

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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

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Correct Option: D
Explanation:

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

If two springs of spring constant $k _1$ and $k _2$ are connected together in parallel, the effective spring constant will be

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

  1. increase

  2. decrease

  3. remain same

  4. none of the above

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Correct Option: A
Explanation:

the effective spring constant = $k _1+k _2$

the correct option is (a)

A block of mass m is suspended separately by two different springs have time period ${ t } _{ 1 }$ and ${ t } _{ 2 }$. If same mass is connected to parallel combination of both springs, then its time period is $T$. Then

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

  1. $T = t _1 + t _2$

  2. $T^2 = t _1^2 + t _2^2$

  3. $T^{-1} = t _1^{-1} + t _2^{-1}$

  4. $T^{-2} = t _1^{-2} + t _2^{-2}$

Show answer
Correct Option: A
Explanation:
Let the spring constant be $k _1$ and $k _2$ 
now, formula of time period of spring is given by 
$T=2\pi \sqrt{m/k}$
now, $t _{1}= 2\pi \sqrt{ m/k _{1}}$
$t _{1}^{2} = \dfrac{4 \pi^{2} m}{k _{1}} \Rightarrow k _{1}=4 \pi^{2} m/t _{1}^{2} $----$(2)$
$t _{2}= 2 \pi \sqrt{m/ k _{}}$
$t _{2}^{2} = \dfrac{4 \pi^{2} m}{k _{2}} \Rightarrow k _{2}= 4 \pi^{2} m/t _{2}^{2}$---- $(2)$
now, both are connected in parallel then ;
$k _{1}+ k _{2} \Rightarrow \dfrac{4 \pi^{2} }{t _{1}^{2}} + \dfrac{4 \pi^{2} m }{t _{2}^{2}}$
$4 \pi^{2} m \left( \dfrac{1}{t _{1}^{2}} + \dfrac{1}{t _{2}^{2}} \right)$
now, time period $=\sqrt{\dfrac{t _{1}^{2}. t _{2}^{2}}{t _{1}^{2}+ t _{2}^{2}}}$

Springs of spring constants K, 2K, 4K, 8K, 2048 K are connected in series. A mass 'm' is attached to one end the system is allowed to oscillation. The time period is approximately :

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

  1. $2\pi \sqrt{\dfrac{m}{2K}}$

  2. $2\pi \sqrt{\dfrac{2m}{2K}}$

  3. $2\pi \sqrt{\dfrac{2m}{K}}$

  4. none of these

Show answer
Correct Option: C
Explanation:
In series the equation spring constant is given by:-
$\dfrac {1}{k'}=\displaystyle \sum \dfrac {1}{k'}\Rightarrow \dfrac {1}{k'}+\dfrac {1}{k}+\dfrac {1}{2k}+\dfrac {1}{4k}+\dfrac {1}{8k}+\dfrac {1}{2048k}$
$\Rightarrow \ \dfrac {1}{k'}=\dfrac {3841}{2048k}$
or approximately $k'\simeq \dfrac {2048k}{3841}\simeq \dfrac {k}{2}$
$\therefore \ $ New time period, $T'=2\pi \sqrt {\dfrac {m}{k'}}=2\pi \sqrt {\dfrac {2m}{k}}$


A body of mass $m$ has time period $T _1$ with one spring and has time period $T _2$ with another spring. if both the spring are connected in parallel and same mass is used, then new time period $T$ is given as

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

  1. $T^{2}= T _{1}^{2}+ T _{2}^{2}$

  2. $T= T _{1}+ T _{2}$

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

  4. $\dfrac{1}{ T^{2}}=\dfrac{1}{ T _{1}^{2}}+\dfrac{1}{ T _{2}^{2}}$

Show answer
Correct Option: D
Explanation:

Angular frequency, $\omega =\sqrt{\dfrac{k}{m}}\ \ \ \Rightarrow \ \dfrac{2\pi }{T}=\sqrt{\dfrac{k}{m}}$

$\Rightarrow \ k=m{{\left( \dfrac{2\pi }{T} \right)}^{2}}$  where, $T$ is time period.

Net Spring constant when two spring is connected in parallel.

$ k={{k} _{1}}+{{k} _{2}} $

$\Rightarrow m{{\left( \dfrac{2\pi }{T} \right)}^{2}}=m{{\left( \dfrac{2\pi }{{{T} _{1}}} \right)}^{2}}+m{{\left( \dfrac{2\pi }{{{T} _{2}}} \right)}^{2}}$

$ \Rightarrow \dfrac{1}{{{T}^{2}}}=\dfrac{1}{{{T} _{1}}^{2}}+\dfrac{1}{{{T} _{2}}^{2}} $ 

When two blocks A and B coupled by a spring on a frictionless table are stretched and then released, then

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

  1. kinetic energy of body at any instatn after releasing is inversely proportional to their masses

  2. kinetic energy of body at any instant may or may not be inversely proportional to their masses

  3. $\cfrac { K.E\quad of\quad B }{ K.E\quad of\quad A } =\cfrac { mass\quad of\quad B }{ mass\quad of\quad A } $

  4. both (b) and (c) are correct

Show answer
Correct Option: A
Explanation:

Force on each block =kx

Let acceleration at any block of mass m is ,
$ma=kx$
$a=\dfrac{kx}{m}$
$\omega=\sqrt{\dfrac{k}{m}}$
velocity $v=at=\dfrac{kx}{m}t$
$KE=\dfrac{1}{2}mv^2$
$=\dfrac{1}{2}m \dfrac{k^2x^2t^2}{m^2}=\dfrac{1}{2}\dfrac{k^2x^2t^2}{m}$
$KE \alpha \dfrac{1}{m}$

Two blocks of masses $m _1$ and $m _2$ are connected by a massless spring and placed on smooth surface. The spring initially stretched and released. Then:

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

  1. The momentum of each particle remains constant seperately

  2. The magnitude of momentums of each body are equal to each other

  3. The mechanical energy of system remains constant

  4. Both (2) &(3)

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Correct Option: D

Two masses $m _{1}=1\ kg$ and $m _{2}=0.5\ kg$ are suspended together by a massless spring of spring constant $12.5\ Nm^{-1}$. When masses are in  equilibrium $m _{1}$ is removed without disturbing the system. New amplitude of oscillation will be 

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

  1. $30\ cm$

  2. $50\ cm$

  3. $80\ cm$

  4. $60\ cm$

Show answer
Correct Option: C
Explanation:
we have formula,

$x _2=\dfrac{m _1g}{k}$

where,

$x _2=amplitude$

$k=spring-constant$

$x _2=\dfrac{1 \times 9.8}{12.5}=0.784\approx 0.8m$

$\therefore x _2=80cm$ is the new amplitude of oscillation.