Tag: horizontal oscillations of a mass attached to a spring

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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

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

  1. $\sqrt{\frac{K}{m}}$
  2. $\sqrt{\frac{m}{K}}$
  3. $\left ( \frac{K}{m} \right )^{2}$
  4. $\left ( \frac{m}{K} \right )^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 )
$\Rightarrow \ w=\sqrt {K/m}$

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

  1. maximum at the extreme position

  2. maximum at the mean position

  3. minimum at the mean position

  4. same at all positions
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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 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

  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. $T^{-2} = t _1^{-2} + t _2^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=\sqrt{\dfrac{t _{1}^{2}. t _{2}^{2}}{t _{1}^{2}+ t _{2}^{2}}}$
Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

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 :

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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, $T'=2\pi \sqrt {\dfrac {m}{k'}}=2\pi \sqrt {\dfrac {2m}{k}}$


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

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

  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. $\dfrac{1}{ T^{2}}=\dfrac{1}{ T _{1}^{2}}+\dfrac{1}{ T _{2}^{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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}} $ 

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $v=at=\dfrac{kx}{m}t$
$KE=\dfrac{1}{2}mv^2$
$=\dfrac{1}{2}m \dfrac{k^2x^2t^2}{m^2}=\dfrac{1}{2}\dfrac{k^2x^2t^2}{m}$
$KE \alpha \dfrac{1}{m}$

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

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:

  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)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a system of two blocks connected by a spring on a smooth surface, the internal spring force acts equally on both, meaning the magnitude of momentum change is equal for both. Since the system is isolated, the total mechanical energy is conserved.

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

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 

  1. $30\ cm$
  2. $50\ cm$
  3. $80\ cm$
  4. $60\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.