Aparticle with mass 1.01 kg oscillates horizontally at the end of a horizontal spring. a student measures an amplitude of 0.923 m and a duration of 125 s for 66 cycles of oscillation. find the frequency, the speed at the equilibrium position, , the spring constant, the potential energy at an endpoint, , the potential energy when the particle is located 56.9% of the amplitude away from the equiliibrium position, and the kinetic energy, k, and the speed, at the same position.

frequency  0.529 Hz

Explanation:

This problem of simple harmonic movement. In this the frequency is the inverse of the period and the period (T) can be calculated as the time it takes for the body to give a certain number of oscillations

    T= t/n

    T = 125/66

    T = 1.89 s

    f =1/T

    f= 0.529 Hz

The expression for the simple harmonic movement is

X = A cos (wt +φ)

To find the speed we derive this formula

V = dx / dt

V = -A w sin (w t + φ)

At the point of equilibrium the function sin is maximum or minimum

Vmax = A w

. w = 2πf

. w = 2π 0.529

. w = 3.32

Vmax = 0.923 3.32

Vmax = 3.06 m / s

The spring constant can be found since the angular velocity is related to the physical constants of the system, spring constant (K) and block mass

. w² = K / m

. K = m w²

. K = 1.01 3.322

. K = 11.1 N / m

For this system the mechanical energy is conserved E = K + U, at the end point the speed is zero, so total mechanical energy is energy potential

E = ½ K A2

E = ½ 11.1 0.9232 = 4.73 J

E = Umax = 4.73 J

At the point at 56.9% of the equilibrium position, for energy is potential and kinetic part, let's start by finding the point value

X = A 56.9 / 100

X = 0.923 56.9 / 100

X = 0.525 m

U = ½ K X²

U = ½ 11.1 0.525²

U = 1.53 J

The kinetic energy can be found from the conservation of mechanical energy E = K + U

 K = E-U

 K = 4.73 -1.53

   K = 3.2 J

To find the velocity we use the equation K = ½ m v2

 . v² = 2K / m

 . v =√ 2 3.2 / 1.01

. v = 2.52 m / s

b) phosphorus pentoxide

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explanation: