Physics

Spring Mass Systems

172 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 physics oscillations introduction to sound free, forced and damped oscillations resonance

Few particles undergo damped harmonic motion. Values for the spring constant $k$ , the damping constant $b$ , and the mass $m$ are given below. Which leads to the smallest rate of loss of mechanical energy at the initial moment?

  1. $ k = 100N/m , m = 50 g, b = 8 g/s $
  2. $ k = 150 N/m , m = 50 g, b = 5 g/s $
  3. $ k = 150N/m , m = 10g, b = 8 g/s $
  4. $ k = 200N/m , m = 8g, b = 6 g/s $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The rate of loss of mechanical energy is proportional to the damping force times velocity, or P = b * v^2. For a given initial displacement, the initial velocity is zero, but the damping force acts as the system moves. The damping coefficient b is the primary factor. Comparing the options, the smallest b value (5 g/s) leads to the smallest rate of energy loss.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

The angular frequency of the damped oscillator is given by $\omega =\sqrt{\left(\frac{k}{m} -\dfrac{r^2}{4m^2}\right)}$ where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio $\dfrac{r^2}{mk}$ is $8%$, the changed in time period compared to the undamped oscillator is approximately as follows:  

  1. Increases by 1%

  2. Decreases by 1%

  3. Decreases by 8%

  4. increases by 8%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\omega = \sqrt {\dfrac{k}{m}-\dfrac{r^2}{4m^2}}= \sqrt{\dfrac{k}{m}}\sqrt{1-\dfrac{r^2}{4mk}}$
 $\approx \omega _o \left(1-\dfrac{r^2}{8mk}\right) \approx$ (1 - 1%)

Multiple choice physics option b: engineering physics introduction to sound free, forced and damped oscillations resonance

In damped oscillation mass is $1\ kg$ and spring constant $=100\ N/m$, damping coefficeint$=0.5\ kg\ s^{-1}$. If the mass displaced by $10\ cm$ from its mean position then what will be the value of its mechanical energy after $4$ seconds?

  1. $0.67\ J$
  2. $0.067\ J$
  3. $6.7\ J$
  4. $0.5\ J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

Mass, $m=1\,kg$

Spring constant, $k=100\,N/m^2$

Damping coefficient, $b=0.5\,kg/s$

Distance, $x=10\,cm$

Time, $t=4\,s$

We know,

The energy for damped oscillation, $E=\dfrac 12kx^2 e^{-\dfrac{bt}{m}}$

$E=\dfrac 12\times 100\times 0.01\times e^{-\dfrac{0.5\times 4}{1}}$

$E=\dfrac{e^{-2}}{2}=0.067\,J$

Hence the mechanical energy is $0.067\,J$
Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

A mass of 50 kg is suspended from a spring of stiffness 10 kN/m. It is set oscillating and it is observed that two successive oscillations have amplitudes of 10 mm and 1 mm. Determine the damping ratio.

  1. 0.315

  2. 0.328

  3. 0.344

  4. 0.353
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


For successive amplitudes $m = 1$
amplitude reduction factor

$=ln\left( \dfrac { { x } _{ 1 } }{ { x } _{ 2 } }  \right) =ln\left(

\dfrac { 10 }{ 1 }  \right) =ln10=2.3$
amplitude reduction factor $=\dfrac { 2\pi \delta m }{ \sqrt { 1-{ \delta  }^{ 2 } }  } $
$\Rightarrow \dfrac { 2\pi \delta m }{ \sqrt { 1-{ \delta  }^{ 2 } }  } =2.3$
squaring both sides
$\dfrac { 39.478{ \delta  }^{ 2 } }{ 1-{ \delta  }^{ 2 } } =5.29\\ \Rightarrow { \delta  }^{ 2 }=0.118\\ \Rightarrow \delta =0.344$

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

A body of mass $\text{600 gm}$ is attached to a spring of spring constant $\text{k = 100 N/m}$ and it is performing damped oscillations.  If damping constant is $0.2$ and driving force is $F = F _{0}$  $cos(\omega t)$  where $F _{0}=20N$  Find the amplitude of oscillation at resonance. 

  1. $\text{4.1 m}$
  2. $\text{0.57 m}$
  3. $\text{7.7 m}$
  4. $\text{0.98 m}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As we know that the amplitude of forced oscillation is given as

$A=\dfrac{F _0}{\sqrt{m^2(\omega^2-\omega _d^2)^2+\omega _d^2b^2}}$

Here we know that when oscillator is in resonance then,

$\omega=\omega _d$

so we have

$A=\dfrac{F _0}{\omega _d b}$

$F _0=20\,N$

$m = 600\, g$

$\omega=\sqrt{\dfrac km}$

$\omega=\sqrt{\dfrac{100}{0.6}}$

$\omega=12.9\,rad/sec$

Now we have

$A=\dfrac{20}{12.9\times 0.2}$

$A=7.7\,m$

Multiple choice physics option b: engineering physics introduction to sound free, forced and damped oscillations resonance

In reality, a spring won't oscillate for ever.               will                the amplitude of oscillation until eventually the system is at rest.

  1. Frictional force, increase

  2. Viscous force, decrease

  3. Frictional force, decrease

  4. Viscous force, increase
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In reality, a spring won't oscillate forever. Frictional force will decrease the oscillation until eventually, the system is at rest.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

Two point masses $m _1$ and $m _2$ are coupled by a spring of spring. Constant $k$ and uncompressed length $L _0$. The spring is fully compressed and a thread ties the masses together with negligible separation between them. The tied assembly is moving in the $+x$ direction with uniform speed $v _0$. At a time, say $t = 0$, it is passing the origin and at that instant the thread breaks. The masses, attached to the spring, start oscillating. The displacement of mass $m _1$ given by $x _1(t) = v _0 t(1 - cos \omega t)$ where $A$ is a constant. Find (i) the displacement $x _2(t)$ is $m _2$, and (ii) the relationship between $A$ and $L _0$.

  1. (i) $v _0 t + \dfrac{m _1}{2m _2}A(1 - cos \omega t)$

    (ii) $A = \left(\dfrac{m _2}{2m _1 + m _2}\right)$
  2. (i) $v _0 t + \dfrac{m _1}{m _2}A(1 - cos \omega t)$

    (ii) $A = \left(\dfrac{m _2}{m _1 + m _2}\right)$
  3. (i) $v _0 t + \dfrac{m _1}{3m _2}A(1 - cos \omega t)$

    (ii) $A = \left(\dfrac{m _2}{3m _1 + m _2}\right)$
  4. (i) $v _0 t + \dfrac{m _1}{4m _2}A(1 - cos \omega t)$

    (ii) $A = \left(\dfrac{m _2}{4m _1 + m _2}\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using conservation of momentum and the properties of a spring-mass system, the center of mass velocity remains constant. The displacement expressions are derived from the relative motion of the two masses oscillating about the center of mass.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

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

  1. 3

  2. 2

  3. 6

  4. 3/2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Specific gravity or relative density of block is given as

$R.D.=\dfrac { mass\quad of\quad block }{ loss\quad in\quad mass\quad in\quad liquid } =\dfrac { 60 }{ 60-40 } $
$\boxed { R.D.=3 } $

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

Two blocks of masses $8$kg are connected by a spring of negligible mass and placed on a frictions less horizontal surface. An impulse gives a velocity of $12$m/s to the heavier block in the direction of lighter block. The velocity of the center of mass is:-

  1. $12$m/s
  2. $10$m/s
  3. $8$m/s
  4. $6$m/s
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Velocity of center of mass is $=\dfrac{m _1v _1+m _2v _2}{m _1+m _2}$


                                                $=\dfrac{8\times 12+8\times 0}{8+8}$


                                                $=\dfrac{96+0}{16}$

                                                $=6m/s$
Hence, the answer is $6m/s.$

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

Two unequal masses are tied together with a cord with a compressed spring in between.
Which one is correct?

  1. Both masses will have equal KE.

  2. Lighter block will have greater KE.

  3. Heavier block will have greater KE.

  4. None of above answers is correct.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Lighter block will have greater kinetic energy to lighter block will have higher velocity mass so Heavier block, hence by equation $\dfrac{1}{2}mv^2,$ the lighter will have greater KE.
Hence, the answer is Lighter block will have greater KE.

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

A spring of spring constant $k$ is cut into $3$ equal part find $k$ of each

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

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The spring constant k is inversely proportional to the length of the spring (k * L = constant). If a spring is cut into 3 equal parts, each part has a length of L/3, so its spring constant becomes 3k.

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

A block of mass m is suddenly released from the top of a string of stiffness constant k.
(i) The maximum compression in the spring will be
(ii) at equilibrium, the compression in the spring will be .......... 

  1. 2mg/k, mg/k

  2. mg/k, mg/k

  3. mg/k, 2mg/k

  4. 2mg/k, 2mg/k
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a mass is released suddenly, the maximum compression is 2mg/k due to energy conservation (potential energy lost equals elastic potential energy gained). At equilibrium, the forces balance (mg = kx), resulting in a compression of mg/k.