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
#SPJ4
