A block whose mass m is 680 g is fastened to a spring whose spring constant k is

Question

A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 N/m
(See figure below). The block is pulled a distance x = 11 cm from its equilibrium position at x = 0.0 on a frictionless surface and released from rest at t = 0.0.
a) What are the angular frequency, the frequency, the amplitude, and the period of the
resulting motion?
b) What is the maximum speed of the oscillating block, and where is the block when it
occurs?
c) What is the magnitude of the maximum acceleration of the block?
d) What is the displacement function x(t) for the spring-block system?

Explanation / Answer

  As F = - k * x is satisfied which is the equation of motion for simple harmonic motion,

   comparing with general equation of motion a = - 2 * x where a = acceleration

      = (k/M)

a)   here k = 65 N/m and m = 680 g. = 9.78 rad/sec

   now frequency ' f ' = / 2 = 1.56 Hz

   time period = 1/ f = 0.643 sec

   amplitude A = maximum displacement = 11 cm

b) the maximum speed of the block will be at equilibrium position

   and maximum speed = A [ from SHM ]

   v (max) = 1.08 m/sec

c) maximum acceleration = A2 [ from SHM ]

   a(max) = 10.52 m/sec2

d) we can write acceleration as

   a = v dv/dx = - kx

   integrating both sides and substituting limits at some intermediate time instance

   velocity = v m/s at displacement = x m and velocity = 0 m/s at disp. = A m.

   we get v2 = [ kA2 - kx2 ] / M

   now substitute v = dx/dt and integrate with required limits we get the differential equation as

   dx / (A2 - x2) = (k/M) dt solving the differential equation we get

   x(t) = A sin(((k/M))t + )

   can be re written as  x(t) = A sin(t + ) where can chosen according to initial conditions.

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