If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced a distance 0.110 m from its equilibrium position and released with zero initial speed. Then after a time 0.815 s its displacement is found to be a distance 0.110 m on the opposite side, and it has passed the equilibrium position once during this interval.
Find (a) the amplitude; (b) the period; (c) the frequency.

Question

contestada

Respuesta :

poojatomarb76 poojatomarb76 · Answer 1 · 2022-04-15T12:28:04+02:00

a)The amplitude of the wave will be 0.120m.

b)The period of the wave will be 1.60s.

Frequency is defined as the number of repetitions of a wave occurring waves in 1 second.

a)The amplitude of the wave will be 0.120m.

A is the initial displacement =0.120m

The amplitude represents the largest deviation from equilibrium. So that the amplitude of the wave will be equal to the maximum position.

Hence the amplitude of the wave will be 0.120m.

b)The period of the wave will be 1.60s.

The movement from maximum positive displacement to maximum negative displacement. The time period will be;

[tex]\rm \frac{T}{2} = 0.800 \\\\\ T= 1.6 \ sec[/tex]

Hence the period of the wave will be 1.60s.

c) The frequency of the wave will be 0.625Hz.

The frequency is inversely proportional to the time period. Frequency is found as;

[tex]\rm f=\frac{1}{T} \\\\ \rm f=\frac{1}{0.625} \\\\ \rm f=0.625 \ Hz[/tex]

Hence the frequency of the wave will be 0.625Hz.

To learn more about the frequency reference the link;

https://brainly.com/question/14926605