0

Horizontal oscillations of a mass attached to a spring - class-XI

Description: horizontal oscillations of a mass attached to a spring
Number of Questions: 47
Created by:
Tags: oscillations simple harmonic motion physics
Attempted 0/47 Correct 0 Score 0

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


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:

  1. Reduce

  2. Reduce impluse

  3. Reduce force

  4. Reduce velocity


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:

  1. maximum at the extreme position

  2. maximum at the mean position

  3. minimum at the mean position

  4. same at all positions


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

  1. increase

  2. decrease

  3. remain same

  4. none of the above


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

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


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 :

  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


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

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


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

  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


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:

  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)


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 

  1. $30\ cm$

  2. $50\ cm$

  3. $80\ cm$

  4. $60\ cm$


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.

A mass of 10g is connected to a massless spring then time period of small oscillation is 10 second. If 10 g mass is replaced by 40 g mass in same spring then its time period will be :-

  1. 5 s

  2. 10 s

  3. 20 s

  4. 40 s


Correct Option: C

Two equal masses are connected by a spring satisfying Hooks law and are placed on a frictionless table. The spring is elongated a little and allowed to go. Let the angular frequency of oscillations be $\omega$. Now one of the masses is stopped. The square of the new angular frequency is :

  1. ${\omega}^{2}$

  2. $\dfrac{{\omega}^{2}}{2}$

  3. $\dfrac{{\omega}^{2}}{3}$

  4. $2{\omega}^{2}$


Correct Option: B

The springs of force constants $K, 2K, 4K, 8K,....128K$ are connected vertically in series and a body of mass M is suspended from the last spring. If this system is set into oscillations, the time period will be- 

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

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

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

  4. $2\pi \sqrt{\dfrac{M}{2K}}$


Correct Option: C

A spring is placed in vertical position by suspending it from a hook at its top. A similar hook on the bottom of the spring is at $11\ cm$ above a table top. A mass of $75\ g$ and of negligible size is then suspended from the bottom hook, which is measured to be $4.5\ cm$ above the table top. The mass is then pulled down a distance of $4\ cm$ and released. Find the approximate position of the bottom hook after $s$?
Take $g=10m/{s}^{2}$ and hooks mass to be negligible.

  1. $5cm$ above the table top

  2. $4.5cm$ above the table top

  3. $9cm$ above the table top

  4. $0.5cm$ above the table top


Correct Option: A

A block of mass m moving with speed v compresses a spring through distance X before its speed is halved.What is the value of spring constant?

  1. $\frac{{3m{v^2}}}{{4{x^2}}}$

  2. $\frac{{m{v^2}}}{{4{x^2}}}$

  3. $\frac{{m{v^2}}}{{2{x^2}}}$

  4. $\frac{{2m{v^2}}}{{{x^2}}}$


Correct Option: A

A spring executes SHM with mass of 10 kg attached to it. The force constant of spring is 10 N/m.If at any instant its velocity is 40 cm/sec. The displacement will be (here amplitude is0.5m) 

  1. 0.06 m

  2. 0.3 m

  3. 0.01 m

  4. 1.0 m


Correct Option: B

Two identical springs are fixed at one end and masses 1$\mathrm { kg }$ and 4$\mathrm { kg }$ are suspended at their other ends. Theyare both stretched down from their mean position and let go simultaneously. If they are in the same phase atter every 4 seconds then the springs constant $\mathrm { k }$ is 

  1. $\pi \frac { N } { m }$

  2. $\pi ^ { 2 } \frac { \mathrm { N } } { \mathrm { m } }$

  3. 2$\pi \frac { \mathrm { N } } { \mathrm { m } }$

  4. given data is insufficient


Correct Option: A

A spring mass system is hanging from the ceiling of an elevator in equilibrium. The elevator suddenly starts accelerating upwards with accelerating a, consider all the options in the reference frame of elevator. 

  1. the frequency of oscillation is $\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { k }{ m } } $

  2. the amplitude of the resulting SHM is $\dfrac { ma }{ k } $

  3. amplitude of resulting SHM is $\dfrac { m\left( g+a \right) }{ k } $

  4. maximum speed of block during oscillation is $\left( \sqrt { \dfrac { m }{ k } } \right) a$


Correct Option: B
Explanation:

Frequency =$2 \pi(m/k)$ frequecy  is independent of $g$ in spring

Extension in spring in equilibrium
initial $ = \dfrac{{mg}}{k}$
Extension in spring in equilibrium  in accelerating lift $m(g+a)$
Amplitude $ = \dfrac{{m\left( {g + a} \right)}}{k} - \dfrac{{mg}}{k} = \dfrac{{ma}}{k}$ 

The reading of a spring balance when a block is suspended from it in air is $60 N$. The reading is changed to $40 N$ when the block is submerged in water. The specific gravity of the block must be therefore.

  1. 3/2

  2. 6

  3. 2

  4. 3


Correct Option: D

A block of mass $1$ kg is connected to a spring of spring constant $\pi^2 N/m$ fixed at other end and kept on smooth level ground. The block is pulled by a distance of $1$ cm from natural length position and released. After what time does the block compress the spring by $\frac{1}{2} cm$.

  1. $\dfrac{2}{3}$ sec

  2. $\dfrac{1}{3}$ sec

  3. $\dfrac{1}{6}$ sec

  4. $\dfrac{1}{12}$ sec


Correct Option: C

A spring (of spring constant $= k )$ is cut into 4 equal parts and two parts are connected in parallel. What is the effective spring constant of these parts?

  1. $4K$

  2. $16K$

  3. $8K$

  4. $6K$


Correct Option: C

A man of mass 50 kg stands on the horizontal platform of a spring balance. The platform starts oscillating with amplitude 0.1 m and frequency $2/  \pi$ Hz. The reading of the spring balance will fluctuate between

  1. 50.8 kg & 49.2 kg

  2. 50 kg & 58 kg

  3. 50 kg & 42 kg

  4. 58 kg & 42 kg


Correct Option: A

A spring of force constant k rests on asmooth floor, with one end fixed to awall. A block of mass $m$ hits the free end of the spring with velocity $v$ . Themaximum force exerted by the springon the wall is 

  1. $v \sqrt { \dfrac {m}{k} }$

  2. $m n \sqrt { k }$

  3. $m \sqrt { ( u k ) }$

  4. $k \sqrt { ( m v ) }$


Correct Option: A

A body of mass 100 gm is suspended from a spring of force constant 50 N/m. The maximum acceleration produced in the spring is:

  1. g/2

  2. g

  3. g/3

  4. g/4


Correct Option: B
Explanation:

To calculate maximum acceleration consider spring at the equilibrium position: 
$F=ma$
where $F$ is force on spring which will be equal to $mg$ only (as gravity is the only force present. 
Therefore, $ma=mg$ or $a=g$.


Two springs mass systems having equal mass and spring constant ${k} _{1}$ and ${k} _{2}$. If the maximum velocities in two systems are equal then ratio of amplitude of 1st so that of 2nd is

  1. $\sqrt{{k} _{1}/{k} _{2}}$

  2. ${k} _{1}/{k} _{2}$

  3. ${k} _{2}/{k} _{1}$

  4. $\sqrt{{k} _{2}/{k} _{1}}$


Correct Option: D

Two bodies A and B of equal mass are suspended from two spearte massless springs of spring contnat ${ K } _{ 1 }$ and  respectively If the bodies oscillate vertically such that their maximum velocities are ewqual , the ratio of the amplitude of A to that of B is 

  1. ${ K } _{ 1 }/{ K } _{ 2 }\quad \quad \quad $

  2. $\quad \sqrt { { K } _{ 1 }/{ K } _{ 2 } } $

  3. $\ { K } _{ 2 }/{ K } _{ 1 }\quad $

  4. $\quad \quad \sqrt { { K } _{ 2 }/{ K } _{ 1 } } \quad \quad \quad $


Correct Option: D

Three masses 700 gm, 500 gm and 400 gm are suspended at the end of the spring and they are in equilibrium. When the 700 gm mass is removed, the system oscillates with a period of 3 sec, when the 500 gm mass is also removed, it will oscillate with a period of 

  1. 1 sec

  2. 2 sec

  3. 3 sec

  4. $\sqrt { \dfrac { 12 }{ 5 } } sec$


Correct Option: B

A uniform spring of force constant $k$ is cut into pieces whose length are in the ratio $1 : 2$. What is the force constant of second piece in terms of $k$?

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

  2. $\dfrac{2k}{3}$

  3. $\dfrac{3k}{2}$

  4. $\dfrac{4k}{3}$


Correct Option: C

The spring constant of a spring is $K$. When it is divided into n equal parts, then what is the spring constant of one piece :

  1. $nK$

  2. $K/n$

  3. $\dfrac{nK}{(n + 1)}$

  4. $\dfrac{(n +1) K}{n}$


Correct Option: A

Two blocks m and m each of mass 3kg is connected with spring of constant 50 N/m. The coefficient of friction between m and ground is 0.4. The maximum amplitude of m during its oscillation, so that m does not move, is 

  1. 24 cm

  2. 12 cm

  3. 2.4 cm

  4. 6 cm


Correct Option: A

A body of mass 'm' when hung from  a long and light spring, the spring stretches by 20 cm The period of vibration of the mass when pulled down the released is 

  1. $\dfrac{2\pi}{7}$

  2. $4$

  3. $ \pi$

  4. $\dfrac{\pi}{\sqrt 3}$


Correct Option: A

A sphere of mass 1 kg is connected to a spring of spring constant $ 5.0 Nm^{-1} $ as shown in figure. A force of 0.5 N is applied on the sphere along X-axis , what is the velocity of the sphere when it is displaced througha distance of 10 cm along X-axis?

  1. $ 0.11 ms^{-1} $

  2. $ 0.22 ms^{-1} $

  3. $ 0.11 cms^{-1} $

  4. $ 0.22 cms^{-1} $


Correct Option: C

A spring of length $'l'$ has spring constant $'k'$ is cut into two parts of length $l _{1}$ and $l _{2}$. If their respective spring constants are $k _{1}$ and $k _{2}$, then $\dfrac {k _{1}}{k _{2}}$ is

  1. $\dfrac {l _{2}}{l _{1}}$

  2. $\dfrac {2l _{2}}{l _{1}}$

  3. $\dfrac {l _{1}}{l _{2}}$

  4. None


Correct Option: A
Explanation:

$k _{1}l _{1} = k _{2}l _{2} = kl$
$\dfrac {k _{1}}{k _{2}} = \dfrac {l _{2}}{l _{1}}$

A block falls from a table $0.6m$ high. It lands on an ideal, mass-less, vertical spring with a force constant of $2.4kN/m$. The spring is initially $25cm$ high, but it is compressed to a minimum height of $10cm$ before the block is stopped. Find the mass of the block $(g=9.81m/s^2)$.

  1. $55.51kg$

  2. $5.51kg$

  3. $0.51kg$

  4. None


Correct Option: B

A block of mass $m=4$ kg undergoes simple harmonic motion with amplitude $A=6$ cm on the frictionless surface. Block is attached to a spring of force constant $k=400 N/m$. If the block is at $x = 6$ cm at time $t = 0$ and equilibrium position is at $x=0$ then the blocks position as a function of time (with $x$ in centimetres and $t$ in seconds)?

  1. $x=6\,sin(10t+\frac{1}{2}\pi)$

  2. $x=6\,sin(10\pi t)$

  3. $x=6\,sin(10\pi t-\frac{1}{2}\pi)$

  4. $x=6\,sin(10t-\frac{1}{4}\pi)$


Correct Option: A
Explanation:


The position of block executing SHM is $x=A\sin\left(\dfrac{{2}{\pi}t}{T}+{\phi}\right)$

where $\phi$ is the initial phase 

given at $t=0$    , $x=6cm$ and $A=6cm$

therefore by above equation $6=6\sin\left(0+{\phi}\right)$

                                        or  $1=\sin\phi$

                                        or $\sin\dfrac{\pi}{2}=\sin\phi$

                                        or $\phi=\dfrac{\pi}{2}$

now time period of system $T=2\pi\sqrt\frac{m}{k}$

                                          $T=2\pi\sqrt\frac{4}{400}$

                                          $T=\pi/5$

so position of block $x=6\sin\left(10t+\dfrac{\pi}{2}\right)$

as the initial phase is in positive x-direction therefore $+ $ sign is taken there.

When a spring-mass system vibrates with simple harmonic motion, the mass in motion reaches its maximum velocity:

  1. when its acceleration is greatest

  2. when its acceleration is least.

  3. once during one oscillation.

  4. when it is as its maximum displacement from equilibrium.

  5. when it experiences maximum force.


Correct Option: B
Explanation:

Velocity in SHM is given by ,

                              $v=\omega\sqrt{a^{2}-y^{2}}$ ,
where $\omega=$ angular velocity ,
           $a=$ amplitude ,
           $y=$ displacement from mean position ,
velocity $v$ will be maximum when RHS of this relation is maximum , this is maximum when $y$ is minimum i.e.$y=0$ ,
now acceleration in SHM is given by ,
                                $A=-\omega^{2}y$ ,
therefore acceleration at $y=0$ will be ,
                                $A=0$ ,
it implies that velocity is maximum when acceleration is least .

A mass of $36$ kg is kept vertically on the top of a massless spring. What is the maximum compression of the spring if the spring constant is $15000 $N/m. Assume $g=10 m/s^2$.
  1. 0.02 m

  2. 0.14m

  3. 0.0004m

  4. 0.2m

  5. 42.52m


Correct Option: A
Explanation:

When the maximum compression of the spring occurs, the forces of gravity and that from spring due to compression balance each other.

At that point, $mg=kx$
$\implies x=\dfrac{mg}{k}$
$=\dfrac{36\times 10}{15000}$
$\approx 0.02m$

A block is attached to an ideal spring undergoes simple harmonic oscillations of amplitude A. Maximum speed of block is calculated at the end of the spring. If the block is replaced by one with twice the mass but the amplitude of its oscillations remains the same, then the maximum speed of the block will

  1. decrease by a factor of 4

  2. decrease by a factor of 2

  3. decrease by a factor of $\sqrt 2$

  4. remain the same

  5. increase by a factor of 2


Correct Option: C
Explanation:

Maximum speed of the block         $V _{max}    = wA$                     where  $w  = \sqrt{\dfrac{k}{m}}$

$\implies$    $V _{max}   = \sqrt{\dfrac{k}{m}} A$                        where $k  = constant$

Now the mass is doubled  i.e  $m' = 2m$  keeping  $A  =constant$ 
$\therefore$       $V' _{max} = \sqrt{\dfrac{k}{2m}} A   = \dfrac{  V _{max}}{\sqrt{2}}$
Thus speed is decreased by a factor of  $\sqrt{2}$.

A block of mass $m$ attached to an ideal spring undergoes simple harmonic motion. The acceleration of the block has its maximum magnitude at the point where :

  1. the speed is the maximum

  2. the potential energy is the minimum

  3. the speed is the minimum

  4. the restoring force is the minimum

  5. the kinetic energy is the maximum


Correct Option: C
Explanation:

The magnitude of acceleration of the simple harmonic motion is $a=F/m=kx/m$

So, the acceleration is maximum when the displacement $x$ from equilibrium will be maximum. Thus, speed will be minimum and also the potential energy will also be maximum and kinetic energy will be minimum at the end points of the oscillation.  

An oscillator consists of a block attached to a spring (k = 400 N/m). At some time t, the position (measured from the system's equilibrium location), velocity and acceleration of the block are x = 0.100m, v = 13.6 m/s, and a = 123 m/s$^2$. The amplitude of the motion and the mass of the block are

  1. $0.2 m, 0.845 kg$

  2. $0.3 m, 0.765 kg$

  3. $0.4 m, 0.325 kg$

  4. $0.5 m, 0.445 kg$


Correct Option: D

A spring balance together with a suspended weight of $2.5$kg is dropped from a height of $30$ metres. The reading on the spring balance, while falling, will show a weight of.

  1. $2.5$kg

  2. $1.25$kg

  3. $0$kg

  4. $25$kg


Correct Option: C
Explanation:

Spring balance reads the net force acting on it by the suspended object. While free falling the object is accelerating with acceleration g m/s-2 . With respect to the spring a pseudo force acts on the object which is equal to mg opposite to the weight of the body. The pseudo force balances the weight of the body and the body does not exert any force on the spring. Thus the spring reading will be zero.

When a Spring of constant K  is cut into 2 equal parts then new spring constant of both the parts would be:

  1. K

  2. 2K

  3. 4K

  4. None of these


Correct Option: B
Explanation:

$F=KL\ K=\cfrac { F }{ L } \ K\propto \cfrac { 1 }{ L } $

So when it is cut into two equal parts its length decreases to half & simultaneously spring constant increases to $2K$

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. $\dfrac { \pi } { 10 } s$


Correct Option: D
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..

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. ${10^4}\ g$


Correct Option: D
Explanation:

$\begin{array}{l} m=0.1\, \, kg,x=9.8\times { 10^{ -2 } }\, \, m,T=6.28\, \, s \ K=\dfrac { { mg } }{ x } \Rightarrow k=10 \ T=2\pi \sqrt { \dfrac { M }{ K }  } \Rightarrow 6.28=2\times 3.14\sqrt { \dfrac { M }{ { 10 } }  }  \ 1=\dfrac { M }{ { 10 } } \Rightarrow M=10\, \, kg={ 10^{ 4 } }g \end{array}$

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. $\displaystyle \sqrt{K _{2}/K _{1}}$


Correct Option: D
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
$V _{1}=V _{2}$
$\sqrt{\dfrac{K _{1}}{m}}A _{1}=\sqrt{\dfrac{K _{2}}{m}}A _{2}$
$\dfrac{A _{1}}{A _{2}}=\sqrt{\dfrac{K _{2}}{K _{1}}}$

In the above question, the velocity of the rear 2 kg block after it separates from the spring will be :

  1. 0 m/s

  2. 5 m/s

  3. 10 m/s

  4. 7.5 m/s


Correct Option: A

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


Correct Option: A
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 $7s$

- Hide questions