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A simple harmonic oscillator has an amplitude of 3. 50 cm and a maximum speed of 26. 0 cm/s. What is its speed when the displacement is 1. 75 cm?.

Sagot :

0.22 m/s is the speed when the displacement is 1.75 cm.

Given:

A simple harmonic oscillator has;

Amplitude (A) =3.50 cm = 0.035 m

Maximum Speed (Vmax) = 26.0 cm/s = 0.26 m/s

Displacment (d) = 1.75cm =0.0175 m

The displacement d, whose maximum is the amplitude A , is expressed as:

∴ d = A Sin wt

   [tex]\frac{d}{A}[/tex] = Sin wt

   t = [tex]\frac{1}{W}[/tex] Sin⁻¹ ( [tex]\frac{d}{A}[/tex] )                  

  v = - Aw cos wt

  v = - Aw cos w [ [tex]\frac{1}{W}[/tex]sin⁻¹ ([tex]\frac{d}{A}[/tex]) ]

  v = - Aw cos [ sin⁻¹ ([tex]\frac{d}{A}[/tex]) ]

Speed, v = Vmax cos [ sin⁻¹ ([tex]\frac{d}{A}[/tex]) ]             ∵ Vmax = Aw

            v = 0.26 cos [ sin⁻¹ ( [tex]\frac{0.0175}{0.035}[/tex]) ]

            v = 0.22 m/s

Therefore, 0.22 m/s is the speed when the displacement is 1.75 cm.

Learn more about simple harmonic motion here:

https://brainly.com/question/20885248

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