52.0k views
2 votes
A block is attached to a spring, with spring constant k, which is attached to a wall. it is initially moved to the left a distance d (at point a and then released from rest, where the block undergoes harmonic motion. the floor is frictionless.

User Tim Bender
by
3.5k points

1 Answer

2 votes

Answer:

x(t) = d*cos ( wt )

w = √(k/m)

Step-by-step explanation:

Given:-

- The mass of block = m

- The spring constant = k

- The initial displacement = xi = d

Find:-

- The expression for displacement (x) as function of time (t).

Solution:-

- Consider the block as system which is initially displaced with amount (x = d) to left and then released from rest over a frictionless surface and undergoes SHM. There is only one force acting on the block i.e restoring force of the spring F = -kx in opposite direction to the motion.

- We apply the Newton's equation of motion in horizontal direction.

F = ma

-kx = ma

-kx = mx''

mx'' + kx = 0

- Solve the Auxiliary equation for the ODE above:

ms^2 + k = 0

s^2 + (k/m) = 0

s = +/- √(k/m) i = +/- w i

- The complementary solution for complex roots is:

x(t) = [ A*cos ( wt ) + B*sin ( wt ) ]

- The given initial conditions are:

x(0) = d

d = [ A*cos ( 0 ) + B*sin ( 0 ) ]

d = A

x'(0) = 0

x'(t) = -Aw*sin (wt) + Bw*cos(wt)

0 = -Aw*sin (0) + Bw*cos(0)

B = 0

- The required displacement-time relationship for SHM:

x(t) = d*cos ( wt )

w = √(k/m)

User Charith Nidarsha
by
3.7k points