Tag: physics

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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 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.

Multiple choice archimedes' principle fluid pressure properties of matter physics

A rectangular block is $10\ cm \times 10\ cm \times 15\ cm$ in size is floating in water with $10\ cm$ side vertical. If it floats with $15\ cm$ side vertical, then the level of water will

  1. $Rise$
  2. $Fall$
  3. $Remain\ same$
  4. $Change\ according\ to\ density\ of\ block$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Archimedes' principle, a floating object displaces a weight of fluid equal to its own weight. Since the weight of the block remains constant regardless of its orientation, the weight (and thus the volume) of the water displaced remains the same, so the water level does not change.

Multiple choice archimedes' principle fluid pressure properties of matter physics

A boat having some iron pieces is floating in a pond. If iron pieces are thrown in the liquid then level of liquid 

  1. Increases

  2. Decreases

  3. May increase or decrease

  4. Neither increases nor decreases
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When the iron pieces are in the boat, they displace a volume of water equal to their weight divided by the density of water. When thrown into the water, they sink and displace only a volume of water equal to their own volume. Since the density of iron is much greater than that of water, the volume displaced when submerged is less than the volume displaced when floating, causing the water level to decrease.

Multiple choice archimedes' principle fluid pressure properties of matter physics

A coil of wire of cross-section$0.50$ $m m ^ { 2 }4$ weighs $75g$ in air and $65g$ in water. The length of the coil in cm is

  1. $\frac { 10 ^ { 2 } } { 50 }$
  2. $\frac { 10 ^ { 2 } } { 0.50 }$
  3. $\frac { 10 ^ { 5 } } { 50 }$
  4. $\frac { 10 ^ { 5 } } { 0.0050 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The loss in weight in water equals the buoyant force, which is the weight of the displaced water. Using density of water (1 g/cm^3), the volume of the coil is 10 cm^3. Given the cross-section of 0.5 mm^2 (0.005 cm^2), the length is Volume/Area = 10 / 0.005 = 2000 cm, which matches 10^5 / 50.