To develop this problem it is necessary to apply the concepts related to speed from the simple harmonic movement. We know that mathematically speed can be expressed as
[tex]v = \omega \sqrt{A^2-x^2}[/tex]
Where,
[tex]\omega[/tex]= Angular velocity
A = Amplitude
x = Displacement
For the expression within the square root to be maximum then x must be equivalent to zero therefore
[tex]v_{max} = A\omega[/tex]
Reorganizing the equation to obtain the highest value of angular velocity we have to
[tex]\omega = \frac{v_{max}}{A}[/tex]
Replacing with our values we have to
[tex]\omega = \frac{26}{3.5}[/tex]
[tex]\omega = 7.426rad/s[/tex]
Finally the speed at the swing point equivalent to 1.75cm would be:
[tex]v = \omega \sqrt{A^2-x^2}[/tex]
[tex]v = (7.428)\sqrt{3.5^2-1.75^2}[/tex]
[tex]v = 22.5cm/s = 0.225m/s[/tex]
Therefore the speed when the displacement of the mass is 1.75cm would be 0.225m/s
