Physics

Spring Mass Systems

170 Questions

A spring mass system is a key physics concept used to study oscillations and simple harmonic motion. It involves understanding spring constants, damping, and series or parallel combinations. These principles are frequently tested in engineering entrance examinations.

Series and parallel springsSpring constant calculationsDamped oscillationsSpring compression energyElevator systems

Spring Mass Systems Questions

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

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. $\dfrac{4k}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a spring of constant k is cut into pieces of length L1 and L2, the constants are k1 = k * (L/L1) and k2 = k * (L/L2). For ratio 1:2, L1 = L/3 and L2 = 2L/3. The second piece (L2) has k2 = k * (L / (2L/3)) = 3k/2.

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

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. $\dfrac{(n +1) K}{n}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a spring is divided into n equal parts, each part has a length L/n. Since k is inversely proportional to length, the new constant k' = k * (L / (L/n)) = n*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' 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. $\dfrac{\pi}{\sqrt 3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The extension x = mg/k = 0.2m. T = 2*pi*sqrt(m/k) = 2*pi*sqrt(x/g) = 2*pi*sqrt(0.2/10) = 2*pi*sqrt(0.02) = 2*pi*sqrt(1/50) = 2*pi / (5*sqrt(2)). This does not match the options. Re-check: if x=20cm=0.2m, T = 2*pi*sqrt(0.2/9.8) = 2*pi*sqrt(1/49) = 2*pi/7.

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

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} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Potential energy lost by block = potential energy gained by spring. m*g*(h + delta_x) = 1/2 * k * (delta_x)^2. h = 0.6m, delta_x = 0.25 - 0.10 = 0.15m. m * 9.81 * (0.6 + 0.15) = 0.5 * 2400 * (0.15)^2. m * 9.81 * 0.75 = 1200 * 0.0225 = 27. m = 27 / (9.81 * 0.75) = 27 / 7.3575 = 3.67 kg. The calculation suggests 5.51 kg may be based on different g or h.

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 $m=4$ 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. $x=6\,sin(10t-\frac{1}{4}\pi)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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 .

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics
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
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $mg=kx$
$\implies x=\dfrac{mg}{k}$
$=\dfrac{36\times 10}{15000}$
$\approx 0.02m$

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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  $\sqrt{2}$.

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 $m$ 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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.  

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

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. $0.5 m, 0.445 kg$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

a = -omega^2 * x. 123 = omega^2 * 0.1. omega^2 = 1230. omega = sqrt(1230) approx 35.07. k = m * omega^2. 400 = m * 1230. m = 400/1230 approx 0.325 kg. v^2 = omega^2 * (A^2 - x^2). (13.6)^2 = 1230 * (A^2 - 0.1^2). 184.96 = 1230 * (A^2 - 0.01). 0.15 = A^2 - 0.01. A^2 = 0.16. A = 0.4 m.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $2K$

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

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. $\dfrac { \pi } { 10 } s$
Reveal answer Fill a bubble to check yourself
D Correct answer
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..