Answer:
66 rpm
Explanation:
The period of oscillation is given by
[tex]T=2\pi \sqrt{\frac {m}{k}}[/tex]
[tex]\frac {k}{m}=\frac {4\pi^{2}}{T^{2}}[/tex] where T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.
Also, net force is given by
Net force=[tex]m\omega^{2} L[/tex]
[tex]\omega=\sqrt{\frac {k\triangle L}{mL}}[/tex] where [tex]\triangle L[/tex] is the elongation, L is original length, [tex]\omega[/tex] is the angular velocity
Substituting the equation of [tex]\frac {k}{m}[/tex] into the above we obtain
[tex]\omega=\sqrt {\frac {4\pi^{2}\triangle L}{T^{2} L}}[/tex]
[tex]\omega=\sqrt {4\pi^{2}\times 0.15L}{0.35^{2}\times L}}=6.952763\approx 6.95 rad/s[/tex]
[tex]6.95\times\frac {60 s}{2\pi rad}\approx 66 rpm[/tex]
