Answer :
The unstretched length of the spring is 3.333 m
Calculation :
k = Spring constant = 1.5 N/m
g = Acceleration due to gravity = 9.81 m/s²
l = Unstretched length
Frequency of SHM motion is given by
[tex]f_{s} = \frac{1}{2\pi }\sqrt{\frac{k}{m} }[/tex]
Frequency of pendulum is given by
[tex]f_{p} = \frac{1}{2\pi }\sqrt{\frac{g}{l} }[/tex]
Given in the question
[tex]f_{s} = \frac{1}{2\pi }\sqrt{\frac{k}{m} }[/tex] = [tex]f_{s} = \frac{1}{2\pi }\sqrt{\frac{g}{l} }[/tex]
[tex]\frac{1}{2\pi }\sqrt{\frac{k}{m} }= \frac{1}{2}\frac{1}{2\pi }\sqrt{\frac{k}{m} }[/tex]
l = [tex]\frac{4gm}{k}[/tex]
l = [tex]\frac{4*9.8*\frac{1.2}{9.8} }{1.44}[/tex]
l = 3.333 m
The unstretched length of the spring is 3.333 m
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