Question

A particle executes simple harmonic motion with an amplitude of 2.00 cm. At what positions does its speed equal one fourth of its maximum speed?

Answer 1

The positions where the speed of a particle executing simple harmonic motion is one fourth of its maximum speed are located at a distance of x = ±X/2 from the equilibrium position.

Step-by-step explanation:

In simple harmonic motion, the speed of a particle is maximum at the equilibrium position and decreases as the particle moves away from the equilibrium. The positions where the speed of the particle is one fourth of its maximum speed are located at a distance of x = ±X/2 from the equilibrium position, where X is the amplitude of the motion.

Since the amplitude of the motion is 2.00 cm, the positions where the speed is one fourth of its maximum speed are at x = ±1.00 cm from the equilibrium position.

Answer 2

The positions are 0.0194 m and - 0.0194 m.

Step-by-step explanation:

Given;

amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m

speed of simple harmonic motion is given as;

`v = \omega √(A^2-x^2)`

the maximum speed of the simple harmonic motion is given as;

`v_(max) = \omega A`

when the speed equal one fourth of its maximum speed

`v =(v_(max))/(4)`

`\omega√(A^2-x^2) = (\omega A)/(4) \\\\√(A^2-x^2)= (A)/(4)\\\\A^2-x^2 = (A^2)/(16) \\\\x^2 = A^2 - (A^2)/(16) \\\\x^2 = (16A^2 - A^2)/(16) \\\\x^2 = (15A^2)/(16) \\\\x= \sqrt{(15A^2)/(16) } \\\\x = \sqrt{(15(0.02)^2)/(16) }\\\\x = 0.0194 \ m \ \ or\ - 0.0194 \ m`

Thus, the positions are 0.0194 m and - 0.0194 m.