Question

A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. What is the period T and frequency f of this oscillation? A mass of 1.2 kg attached to a spring is in simple harmonic motion along the x-axis with a period T = 2.5 s. If the total energy of the system is 2.7 J, what is the amplitude of the oscillation? An 8.0-kg block is attached to a spring with a spring constant of . If the spring is stretched 3.0 cm from its equilibrium position and released from rest, the maxium velocity attained by the mass is? Enter question here...

Answer 1

Answer:

a)  T = 1,467 s , b)    A = 0.495 m , c)  v = 4.97 10⁻² m / s

Explanation:

The simple harmonic movement is described by the expression

        x = A cos (wt + Ф)

Where the angular velocity is

       w = √ k / m

a) Ask the period

Angular velocity, frequency and period are related

      w = 2π f = 2π / T

      T = 2π / w

      T = 2pi √ m / k

      T = 2π √ (1.2 / 22)

      T = 1,467 s

      f = 1 / T

      f = 0.68 Hz

b) ask the amplitude

The mechanical energy of a harmonic oscillator

        E = ½ k A²

       A = √2 E / k

       A = √ (2 2.7 / 22)

       A = 0.495 m

c) the mass changes to 8.0 kg

As released from rest Ф = 0, the equation remains

         x = A cos wt

        w = √ (22/8)

        w = 1,658

         x = 3.0 cos (1,658 t)

Speed ​​is

         v = dx / dt

         v = -A w sin wt

The speed is maximum when without wt = ±1

         v = Aw

         v = 0.03    1,658

         v = 4.97 10⁻² m / s