For a simple harmonic motion with given angular frequency `omega`, two arbitrary initial conditions are necessary and sufficient to determine the motion completely. These initial conditions may beA. initial position and initial velocityB. amplitude and initial phaseC. total energy of oscillation and amplitudeD. total energy of oscillation and initial phase.

For a simple harmonic motion with given angular frequency `omega`, two

1.	For a simple harmonic motion with given angular frequency `omega`, two arbitrary initial conditions are necessary and sufficient to determine the motion completely. These initial conditions may beA. initial position and initial velocityB. amplitude and initial phaseC. total energy of oscillation and amplitudeD. total energy of oscillation and initial phase.
Answer» Correct Answer - A::B::D Description of motion is completely specified if we know the variation of x as a function of time. For simple harmonic motion, the general equation of motion is `x=A(omegat+delta)`. As `omega` is given to describe the motion completely, we need the values of A and `delta`. From options (b) and (d) we can have the value of A and `delta` derectly. For option (a) we can find A and `delta` if we know initial velocity and initial position. Option (c ) cannot give the value of A and `delta` so it is not the correct condition.

Discussion

No Comment Found

Related InterviewSolutions

Two identical blocks A and B, each of mass `m=3kg`, are connected with the help of an ideal spring and placed on a smooth horizontal surface as shown in Fig. Another identical blocks C moving velocity `v_0=0.6(m)/(s)` collides with A and sticks to it, as a result, the motion of system takes place in some way Based on this information answer the following questions: Q. After the collision of C and A, the combined body and block B wouldA. oscillate about centre of mass of system and centre of mass is at rest.B. oscillate about centre of mass of system and centre of mass is moving.C. oscillate but about different location other than the centre of mass.D. not oscillate.
The potential energy of a particle of mass `1kg` in motion along the x- axis is given by: `U = 4(1 - cos 2x)`, where `x` in meters. The period of small oscillation (in sec) isA. `2pi`B. `pi`C. `(pi)/(2)`D. `sqrt2pi`
A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second it travels a distance b in the same direction. The amplitude of the SHM isA. `(2a^2)/(3b-a)`B. `(3a^2)/(3a-b)`C. `(2a^2)/(3a-b)`D. `(3a^2)/(3b-a)`
Assume that a tunnel is dug across the earth (radius=R) passing through its centre. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of `sqrt((gR)` (b) it is relased from a height R above the tunnel (c ) it is thrown vertically upward along the length of tunnel with a speed of `sqrt(gR).`
A particle free to move along the (x - axis) hsd potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, where (k) is a positive constant of appropriate dimensions. Then.A. for small displacement from `x=0`, the motion is simple harmonic.B. if its total mechanical energy is `(k)/(2)`, it has its minimum kinetic energy at the originC. for any finite non zero value of x, there is a force directed away from the originD.
In the previous question, the magnitude of velocity of particle at the mentioned instant isA. `(piA)/(T)`B. `(sqrt2piA)/(T)`C. zeroD. `sqrt((7)/(8))xx(2piA)/(T)`
In the previous problem, the displacement of the particle from the mean position corresponding to the instant mentioned isA. `(5)/(pi)m`B. `(5sqrt3)/(pi)m`C. `(10sqrt3)/(pi)m`D. `(5sqrt3)/(2pi)m`
In the previous question, find maximum velocity and maximum acceleration.
The oscillations represented by curve 1 in the graph are expressed by equation`x=Asinomegat.` The equation for the oscillation represented by curve `2` is expressed asA. `x=2Asin(omegat-(pi)/(2))`B. `x=2Asin(omegat+(pi)/(2))`C. `x=-2Asin(omegat-(pi)/(2))`D. `x=Asin(omegat-(pi)/(2))`
A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. Propertional to `(1)/(sqrta)`B. Independent of a.C. Propertional to `sqrta`D. Propertional to `a^(3//2)`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment