Tag: resonance

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

Undamped oscillations are practically impossible because

  1. there is always loss of energy.

  2. there is no force opposing friction.

  3. energy is not conserved in such oscillations.

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

Underdamped oscillation is practical because there always be resistive force present in reality which will try to make an oscillating body to lose its energy. This loss of energy makes the motion damped motion.

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

If we wish to represent the equation for the position of the mass in terms of a differential equation, which one of these would be the most suitable?

  1. $ m \dfrac{d^2x}{dt^2} + b \dfrac{dx}{dt} + kx = 0$
  2. $ m \dfrac{d^2x}{dt^2} - b \dfrac{dx}{dt} + kx = 0$
  3. $ m \dfrac{d^2x}{dt^2} + b \dfrac{dx}{dt} - kx = 0$
  4. $ m \dfrac{d^2x}{dt^2} -b \dfrac{dx}{dt} - kx = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The force on body oscillating in resistive medium is 

$f = -kx - bv$
$\Rightarrow m\dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } =-kx-b\dfrac { dx }{ dt } \ \Rightarrow m\dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +b\dfrac { dx }{ dt } +kx=0$
k = oscillating constant 
x = displacement of body from mean position 
b = constant depends on resistive medium 
v = velocity of object = $\dfrac{dx}{dt}$
m = mass of object .

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.