The amplitude of vibration of a particle is given by `a_m=(a_0)//(aomega^2-bomega+c)`, where `a_0`, a, b and c are positive. The condition for a single resonant frequency is
A
`b^(2) = 4ac`
B
`b^(2) gt 4ac`
C
`b^(2) = 5ac`
D
`b^(2)= 7ac`
Text Solution
Verified by Experts
The correct Answer is:
A
`a omega^(2) - b omega +c = 0`
Topper's Solved these Questions
OSCILLATIONS
NARAYNA|Exercise LEVEL-II (H.W)|1 Videos
OSCILLATIONS
NARAYNA|Exercise DISPLACEMENT, VELOCITY AND ACCELERATION|13 Videos
OSCILLATIONS
NARAYNA|Exercise SIMPLE PENDULUM|14 Videos
NEWTONS LAWS OF MOTION
NARAYNA|Exercise PASSAGE TYPE QUESTION|6 Videos
PHYSICAL WORLD
NARAYNA|Exercise C.U.Q|10 Videos
Similar Questions
Explore conceptually related problems
The amplitude of velocity of a particle is given by, V_(m)=V_(0)//(a omega^(2)-b omega+c) where V_(0) , a, b and c are positive : The condition for a single resonant frequency is
The velocity of the particle of mass m as a function of time t is given by v = Aomega.cos[sqrt(K/m)t] , where A is amplitude of oscillation. The dimension of A/K is
The period of oscillation of a particle in SHM is 4sec and its amplitude of vibration is 4cm . The distance of the particle 0.5s after passsing the mean position is
The first obrital of H or H like atom is represencted by psi = 1/( sqrt pi) (Z/a_0) ^(3//2) e^(-ze//a_0) where a_0 = Bohr's orbit . The actual probability of fiding the elercrton at a distance r form the nucleus is :
The cubical container ABCDEFGH which is completely filled with an ideal (nonviscous and incompressible) fluid, moves in a gravity free space with a acceleration of a=a_0 (hat i -hatj +hat k) where a_0 is a positive constant. Then the minimum pressure at the point will be
A particle of mass m moves under the action of a central force. The potential energy function is given by U(r)=mkr^(3) Where k is a positive constant and r is distance of the particle from the centre of attraction. (a) What should be the kinetic energy of the particle so that it moves in a circle of radius a0 about the centre of attraction? (b) What is the period of this circualr motion ?
On the superposition of the two waves given as y_1=A_0 sin( omegat-kx) and y_2=A_0 cos ( omega t -kx+(pi)/6) , the resultant amplitude of oscillations will be
The wave function of 2s electron is given by W_(2s) = (1)/(4sqrt(2pi))((1)/(a_(0)))^(3//2)(2 - (r )/(a_(0)))e^(-1 a0) It has a node at r = r_(p) .Find the radiation between r_(p) and a
NARAYNA-OSCILLATIONS-DAMPED AND FORCED OSCILLATIONS