Answer:
2.67kg
Explanation:
The maximum velocity, [tex]v _ {max}[/tex] of a body experiencing simple harmonic motion is given by equation (1);
[tex]v_{max}=\omega A............(1)[/tex]
where [tex]\omega[/tex] is the angular velocity and A is the amplitude.
The problem describes the oscillation of a loaded spring, and for a loaded spring the angular velocity is given by equation (2);
[tex]\omega=\sqrt{\frac{k}{m}}.................(2)[/tex]
where k is the force constant of the spring and m is the loaded mass.
We can make [tex]\omega[/tex] the subject of formula in equation (1) as follows;
[tex]\omega=\frac{v_{max}}{A}.................(3)[/tex]
We then combine equations (2) and (3) as follows;
[tex]\frac{v_{max}}{A}=\sqrt{\frac{k}{m}}.................(4)[/tex]
According to the problem, the following are given;
[tex]v_ {max }=1.92m/s\\A=0.21m\\k=223N/m[/tex]
We then substitute these values into equation (4) and solve for the unknown mass m as follows;
[tex]\frac{1.92}{0.21}=\sqrt{\frac{223}{m}}[/tex]
[tex]9.143=\sqrt{\frac{223}{m}}[/tex]
Squaring both sides, we obtain the following;
[tex]9.143^2=\frac{223}{m}\\9.143^2*m=223\\83.592m=223\\therefore\\m=\frac{223}{83.592}\\m=2.67kg[/tex]
