Question

Question 1 1 21 Define simple harmonic motion (SHM). a) b) A steel rule is clamped to the end of a table, so that a large section overhangs the edge of the table. The free end of the rule vibrates with SHM when it is pulled down 30.5 mm and released. The frequency of the vibration is 16.0 Hz. i. Determine the time period of the vibration. I 2 1 i. Determine the maximum speed of the free end of the rule during the 3 1 vibration ii. Determine the magnitude of the acceleration when the end of the 121 rule is 20.4 mm from its equilibrium position. iv. Determine the time taken for the free end of the rule to move directly I5 from a position 15.0 mm above the equilibrium position to 15.0 mm below the equilibrium position. v. Some fine particles of sand are sprinkled on the free end of the rule. [3 Determine the maximum amplitude of the vibration if the particles are to remain in contact with the rule throughout the vibration. vi. Draw a diagram to show how the potential, kinetic and total energy 31 of the sand particles will vary with displacement over one cycle of vibration, assuming that they remain in contact with the rule.

Answer 1

a) A simple harmonic motion is one in which the acceleration is directly proportional to the displacement of the object undergoing motion and the direction of the acceleration is opposite to the displacement. Eg: Simple Pendulum or circular motion.

b)

i) Time period = T = 1/frequency = 0.0625 s

ii) maximum speed = amplitude x Angular frequency

$v = A\omega = A(2\pi f) = 3.066 m/s$

iii) for a simple harmonic motion, the magnitude of acceleration is given by:

$a = \omega^2x$

Where x is the displacement from the equilibrium position.

For x = 20.4 mm,

.

(16))2(20.4 × 10-3) = 206 . 17m/s- a = (2

iv) Reaching the opposite end of the motion (from 15mm above to 15mm below) will take one-half of the time period of the motion. So, the time it takes to do this is:

t = T/2 = 0.03175s.