Advertisements
Advertisements
प्रश्न
A particle is performing simple harmonic motion with amplitude A and angular velocity ω. The ratio of maximum velocity to maximum acceleration is ______.
विकल्प
ω
1/ω
ω2
A/ω
Advertisements
उत्तर
A particle is performing simple harmonic motion with amplitude A and angular velocity ω. The ratio of maximum velocity to maximum acceleration is 1/ω.
APPEARS IN
संबंधित प्रश्न
Choose the correct option:
The graph shows variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph?
Answer in brief.
Using differential equations of linear S.H.M, obtain the expression for (a) velocity in S.H.M., (b) acceleration in S.H.M.
Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75 m/s2 to 9.8 m/s2.
A particle is performing S.H.M. of amplitude 5 cm and period of 2s. Find the speed of the particle at a point where its acceleration is half of its maximum value.
Using the differential equation of linear S.H.M., obtain an expression for acceleration, velocity, and displacement of simple harmonic motion.
In U.C.M., when time interval δt → 0, the angle between change in velocity (δv) and linear velocity (v) will be ______.
A wheel of M.I. 50 kg m2 starts rotating on applying a constant torque of 200 Nm. Its angular velocity after 2.5 second from the start is ______.
A particle is moving along a circular path of radius 6 m with a uniform speed of 8 m/s. The average acceleration when the particle completes one-half of the revolution is ______.
If 'α' and 'β' are the maximum velocity and maximum acceleration respectively, of a particle performing linear simple harmonic motion, then the path length of the particle is _______.
A particle executing S.H.M. has amplitude 0.01 m and frequency 60 Hz. The maximum acceleration of the particle is ____________.
A body of mass 5 g is in S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/s. Its velocity will be 50 cm/s at a distance of, ____________.
The displacement of a particle is 'y' = 2 sin `[(pit)/2 + phi]`, where 'y' is cm and 't' in second. What is the maximum acceleration of the particle executing simple harmonic motion?
(Φ = phase difference)
The length of the second's pendulum is decreased by 0.3 cm when it is shifted from place A to place B. If the acceleration due to gravity at place A is 981 cm/s2, the acceleration due to gravity at place B is ______ (Take π2 = 10)
A body is executing S.H.M. Its potential energy is E1 and E2 at displacements x and y respectively. The potential energy at displacement (x + y) is ______.
The bob of a simple pendulum is released at time t = 0 from a position of small angular displacement. Its linear displacement is ______.
(l = length of simple pendulum and g = acceleration due to gravity, A = amplitude of S.H.M.)
A block of mass 16 kg moving with velocity 4 m/s on a frictionless surface compresses an ideal spring and comes to rest. If force constant of the spring is 100 N/m then how much will be the spring compressed?
The displacement of the particle performing S.H.M. is given by x = 4 sin πt, where x is in cm and t is in second. The time taken by the particle in second to move from the equilibrium position to the position of half the maximum displacement, is ______.
`[sin30^circ=cos60^circ=0.5, cos30^circ=sin60^circ=sqrt3/2]`
The displacement of a particle in S.H.M. is x = A cos `(omegat+pi/6).` Its speed will be maximum at time ______.
The displacements of two particles executing simple harmonic motion are represented as y1 = 2 sin (10t + θ) and y2 = 3 cos 10t. The phase difference between the velocities of these waves is ______.
A particle is performing SHM starting extreme position, graphical representation shows that between displacement and acceleration there is a phase difference of ______.
A spring of force constant of 400 N/m is loaded with a mass of 0.25 kg. The amplitude of oscillations is 4 cm. When mass comes to the equilibrium position. Its velocity is ______.
A body of mass 0.5 kg travels in a straight line with velocity v = ax3/2 where a = 5 m–1/2s–1. The change in kinetic energy during its displacement from x = 0 to x = 2 m is ______.
In the given figure, a = 15 m/s2 represents the total acceleration of a particle moving in the clockwise direction on a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is ______.
Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by x = `5sin((pit)/3)`m.
State the expression for the total energy of SHM in terms of acceleration.
A particle executing SHM has velocities v1 and v2 when it is at distance x1 and x2 from the centre of the path. Show that the time period is given by `T=2pisqrt((x_2^2-x_1^2)/(v_1^2-v_2^2))`
