हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

Motion of an oscillating liquid column in a U-tube is ______.

Advertisements
Advertisements

प्रश्न

Motion of an oscillating liquid column in a U-tube is ______.

विकल्प

  • periodic but not simple harmonic.

  • non-periodic.

  • simple harmonic and time period is independent of the density of the liquid.

  • simple harmonic and time-period is directly proportional to the density of the liquid.

MCQ
रिक्त स्थान भरें
Advertisements

उत्तर

Motion of an oscillating liquid column in a U-tube is simple harmonic and time period is independent of the density of the liquid.

Explanation:

If the liquid in U-tube is filled to a height of h and the cross-section of the tube is uniform and the liquid is incompressible and non-viscous. Initially, the level of liquid in the two limbs will be at the same height equal to h. If the liquid is pressed by y in one limb, it will rise by y along the length of the tube in the other limb, so the restoring force will be developed by the hydrostatic pressure difference

Here, the hydrostatic pressure provides the restoring force, thus,

F = – V.p.g

= – A.2ypg, where,

A = Area of a cross-section of tube and F ∝ – y,

Thus, it is a simple harmonic motion.

T = `sqrt((2πm("inertia"))/(k("spring"))`

= `sqrt((2πA(2h))/(p2Apg))`

T = `2π sqrt(h/g)`

Thus, the motion is harmonic as the time period is independent of density.

shaalaa.com
Energy in Simple Harmonic Motion
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 14: Oscillations - Exercises [पृष्ठ ९८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 11
अध्याय 14 Oscillations
Exercises | Q 14.4 | पृष्ठ ९८

संबंधित प्रश्न

A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?


The equation of motion of a particle started at t = 0 is given by x = 5 sin (20t + π/3), where x is in centimetre and t in second. When does the particle
(a) first come to rest
(b) first have zero acceleration
(c) first have maximum speed?


Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos (50 πt + tan−1 0.75) where x is in centimetre and t in second. The motion is started at t = 0. (a) When does the particle come to rest for the first time? (b) When does he acceleration have its maximum magnitude for the first time? (c) When does the particle come to rest for the second time ?


A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum, i.e., a pendulum having frequency same as that of the block.


A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4 s. If the oscillations are stopped and the body hangs in equilibrium find the potential energy stored in the spring.


The block of mass m1 shown in figure is fastened to the spring and the block of mass m2 is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance (2/k) (m1 + m2)g sin θ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of blocks at the time of separation?


The spring shown in figure is unstretched when a man starts pulling on the cord. The mass of the block is M. If the man exerts a constant force F, find (a) the amplitude and the time period of the motion of the block, (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at this position.


The springs shown in the figure are all unstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.


Find the elastic potential energy stored in each spring shown in figure, when the block is in equilibrium. Also find the time period of vertical oscillation of the block.


Show that for a particle executing simple harmonic motion.

  1. the average value of kinetic energy is equal to the average value of potential energy.
  2. average potential energy = average kinetic energy = `1/2` (total energy)

Hint: average kinetic energy = <kinetic energy> = `1/"T" int_0^"T" ("Kinetic energy") "dt"` and

average potential energy = <potential energy> = `1/"T" int_0^"T" ("Potential energy") "dt"`


When a particle executing S.H.M oscillates with a frequency v, then the kinetic energy of the particle? 


When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is ______.


A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is ______.


A body is performing S.H.M. Then its ______.

  1. average total energy per cycle is equal to its maximum kinetic energy.
  2. average kinetic energy per cycle is equal to half of its maximum kinetic energy.
  3. mean velocity over a complete cycle is equal to `2/π` times of its π maximum velocity. 
  4. root mean square velocity is times of its maximum velocity `1/sqrt(2)`.

Displacement versus time curve for a particle executing S.H.M. is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.


Draw a graph to show the variation of P.E., K.E. and total energy of a simple harmonic oscillator with displacement.


The total energy of a particle, executing simple harmonic motion is ______.

where x is the displacement from the mean position, hence total energy is independent of x.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×