Advertisements
Advertisements
Question
Motion of an oscillating liquid column in a U-tube is ______.
Options
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.
Advertisements
Solution
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.
APPEARS IN
RELATED QUESTIONS
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?
A particle having mass 10 g oscillates according to the equation x = (2.0 cm) sin [(100 s−1)t + π/6]. Find (a) the amplitude, the time period and the spring constant. (c) the position, the velocity and the acceleration at t = 0.
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?
The pendulum of a clock is replaced by a spring-mass system with the spring having spring constant 0.1 N/m. What mass should be attached to the spring?
A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.
In following figure k = 100 N/m M = 1 kg and F = 10 N.
- Find the compression of the spring in the equilibrium position.
- A sharp blow by some external agent imparts a speed of 2 m/s to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at this instant.
- Find the time period of the resulting simple harmonic motion.
- Find the amplitude.
- Write the potential energy of the spring when the block is at the left extreme.
- Write the potential energy of the spring when the block is at the right extreme.
The answer of b, e and f are different. Explain why this does not violate the principle of conservation of energy.
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.
Repeat the previous exercise if the angle between each pair of springs is 120° initially.
Consider the situation shown in figure . Show that if the blocks are displaced slightly in opposite direction and released, they will execute simple harmonic motion. Calculate the time period.
A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.
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.
- the average value of kinetic energy is equal to the average value of potential energy.
- 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"`
A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is ______.
A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is ______.
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.
A mass of 2 kg is attached to the spring of spring constant 50 Nm–1. The block is pulled to a distance of 5 cm from its equilibrium position at x = 0 on a horizontal frictionless surface from rest at t = 0. Write the expression for its displacement at anytime t.
An object of mass 0.5 kg is executing a simple Harmonic motion. Its amplitude is 5 cm and the time period (T) is 0.2 s. What will be the potential energy of the object at an instant t = `T/4` s starting from the mean position? Assume that the initial phase of the oscillation is zero.
