Answer: 1.39 s
Explanation:
We can solve this problem with the following equations:
[tex]\frac{\Delta l}{l_{o}}=\frac{F}{AY}[/tex] (1)
[tex]T=2 \pi \sqrt{\frac{l_{o}}{g}}[/tex] (2)
Where:
[tex]\Delta l=0.05 mm=5(10)^{-5} m[/tex] is the length the steel wire streches (taking into account 1mm=0.001 m)
[tex]l_{o}[/tex] is the length of the steel wire before being streched
[tex]F=mg=(2 kg)(9.8 m/s^{2})=19.6 N[/tex] is the force due gravity (the weight) acting on the pendulum with mass [tex]m=2 kg[/tex]
[tex]A[/tex] is the transversal area of the wire
[tex]Y=2(10)^{11} Pa[/tex] is the Young modulus for steel
[tex]T[/tex] is the period of the pendulum
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
Knowing this, let's begin by finding [tex]A[/tex]:
[tex]A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}[/tex] (3)
Where [tex]d=1.1 mm=0.0011 m[/tex] is the diameter of the wire
[tex]A=\pi \frac{(0.0011 m)^{2}}{4}[/tex] (4)
[tex]A=9.5(10)^{-7}m^{2}[/tex] (5)
Knowing this area we can isolate [tex]l_{o}[/tex] from (1):
[tex]l_{o}=\frac{\Delta l AY}{F}[/tex] (6)
And substitute [tex]l_{o}[/tex] in (2):
[tex]T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}[/tex] (7)
[tex]T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}[/tex] (8)
Finally:
[tex]T=1.39 s[/tex]
