Question

S A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a frictionless, horizontal track (Fig. P15.75). The force constant of the spring is k, and the equilibrium length is l . Assume all portions of the spring oscillate in phase and the velocity of a segment of the spring of length d x is proportional to the distance x from the fixed end; that is, vx = (x / l) v . Also, notice that the mass of a segment of the spring is dm = (m /l) dx . Find (b) the period of oscillation.

Answer 1

The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

What is period of oscillation?

This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.

The period of oscillation is given as T

T = 2 * pi * sqrt ( m / k )

where

m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:

m1 = mass of the block

m2 = mass pf the spring

k = force constant of the spring

including the two masses to the period gives

T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

Read more on period of oscillation here: brainly.com/question/22499336

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