Question

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?.

Answer 1

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

   $\frac{d}{A}$ = Sin wt

   t = $\frac{1}{W}$ Sin⁻¹ ( $\frac{d}{A}$ )                  

  v = - Aw cos wt

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

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

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

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

            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:

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