Advertisements
Advertisements
प्रश्न
Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by x = `5sin((pit)/3)`m.
Advertisements
उत्तर १
Given:
x = `5sin((pit)/3)`m
t = 1s
∴ ν = `(dx)/(dt) = d/dt(5sin((pit)/3)) = 5cos((pit)/3) xx pi/3`
Put t = 1s ...(Given)
∴ ν = `5cos(pi/3) xx pi/3` = 2.6179 m/s
उत्तर २
t = 1s, x = 5 sin60
A = 5, x = `2.5sqrt3` ...........(given)
v = `wsqrt(A^2 - x^2)` .........(Formula)
v = `pi/3sqrt(25 - 18.74)`
v = 1.04 × 2.5
v = 2.61 m/s
APPEARS IN
संबंधित प्रश्न
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 simple harmonic motion with amplitude A and angular velocity ω. The ratio of maximum velocity to maximum acceleration is ______.
Using the differential equation of linear S.H.M., obtain an expression for acceleration, velocity, and displacement of simple harmonic motion.
Two identical wires of substances 'P' and 'Q ' are subjected to equal stretching force along the length. If the elongation of 'Q' is more than that of 'P', then ______.
In U.C.M., when time interval δt → 0, the angle between change in velocity (δv) and linear velocity (v) will be ______.
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 ______.
A body performing a simple harmonic motion has potential energy 'P1' at displacement 'x1' Its potential energy is 'P2' at displacement 'x2'. The potential energy 'P' at displacement (x1 + x2) 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 _______.
Which of the following represents the acceleration versus displacement graph of SHM?
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 simple pendulum of length 'L' is suspended from a roof of a trolley. A trolley moves in horizontal direction with an acceleration 'a'. What would be the period of oscillation of a simple pendulum?
(g is acceleration due to gravity)
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 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 body perform linear simple harmonic motion of amplitude 'A'. At what displacement from the mean position, the potential energy of the body is one fourth of its total energy?
A particle is performing SHM starting extreme position, graphical representation shows that between displacement and acceleration there is a phase difference of ______.
In figure, a particle is placed at the highest point A of a smooth sphere of radius r. It is given slight push and it leaves the sphere at B, at a depth h vertically below A, such that h is equal to ______.
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 particle of mass 5 kg moves in a circle of radius 20 cm. Its linear speed at a time t is given by v = 4t, t is in the second and v is in ms-1. Find the net force acting on the particle at t = 0.5 s.
The displacement of a particle of mass 3 g executing simple harmonic motion is given by Y = 3 sin (0.2 t) in SI units. The kinetic energy of the particle at a point which is at a distance equal to `1/3` of its amplitude from its mean position 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 ______.
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))`
