A particle executes linear simple harmonic motion withfrequency 0.35 Hz about th

Question

A particle executes linear simple harmonic motion withfrequency 0.35 Hz about the point x = 0. At t=0, it hasdisplacement x = 0.33cm and zero velocity. Determine thefollowing: using a. the period b. the angular frequency c. the amplitude d. the maximum speed e. the magnitude of the maximum acceleration A particle executes linear simple harmonic motion withfrequency 0.35 Hz about the point x = 0. At t=0, it hasdisplacement x = 0.33cm and zero velocity. Determine thefollowing: using a. the period b. the angular frequency c. the amplitude d. the maximum speed e. the magnitude of the maximum acceleration

Explanation / Answer

The equation for displacement: y=Asin(t+) Eq for velocity= Acos(t+) f=0.35hz a) T=1/f     =2.86s b) =2f         =2.2rad/sc) when t=0, Asin=0.33 and Acos=0 we know that A and are positive, =90degrees. Asin90=0.33 A=0.33cm d)v(max)=A (A in meters)              =0.0033*2.2              =0.00726 m/s OR 0.726 cm/s e) a(max)=-A2 = -0.0033*2.22                =0.0159 m/s2 OR 1.59cm/s2

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