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A block of mass m is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring. The spring has force constant k, and the amplitude of the SHM is A. The block has v = 0, and x = +A at t = 0. It first reaches x = 0 when t=T/4, where T is the period of the motion.

a) In terms of T, what is the time t when the block first reaches x=A/2?

b)The block has its maximum speed when t=T/4. What is the value of t when the speed of the block first reaches the value vmax/2?

c)(Hidden. will ask in comments)

Respuesta :

Answer

We know equation of motion

     x = A cos ω t

a) time t when x = A/2

   we know

  x = A cos ω t

  A/2 = A cos ω t

  cos ω t = 0.5

   ω t = π/3

  [tex]\dfrac{2\pi}{T}\times t = \dfrac{\pi}{3}[/tex]

  [tex]t = \dfrac{T}{6}[/tex]

B) velocity

 [tex]v =\dfrac{dx}{dt} =- A \omega sin\omega t[/tex]

speed becomes half which means

  sin ω t = 1/2

  [tex]\dfrac{2\pi}{T}\times t = \dfrac{\pi}{6}[/tex]

  [tex]t = \dfrac{T}{12}[/tex]

onyebuchinnaji

The time t when the block first reaches half the amplitude is [tex]\frac{T}{6}[/tex].

The value of t when the speed of the block first reaches half of its maximum value is [tex]\frac{T}{12}[/tex].

The given parameters:

  • Spring constant = k
  • Amplitude of the oscillation, = A
  • Period of oscillation, = T

The general wave equation is given as;

x = A cos ω t

The time when the block reaches x = A/2

[tex]\frac{A}{2}= A \ cos \ \omega t\\\\\frac{1}{2} = cos \ \omega t\\\\0.5 = cos \ \omega t\\\\\omega t = cos^{-1} (0.5)\\\\\omega t = \frac{\pi}{3} \\\\(\frac{2\pi }{T} )t = \frac{\pi}{3}\\\\t = \frac{T}{6}[/tex]

The value of t when the speed of the block first reaches half of its maximum value;

[tex]v_{max} = sin \ \omega t\\\\\frac{1}{2} = sin \ \omega t\\\\\frac{\pi}{6} = \omega \ t\\\\\frac{\pi}{6} = (\frac{2\pi }{T} ) \times t\\\\\frac{T}{12} = t[/tex]

Learn more about period of SMH here: https://brainly.com/question/16968916

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