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Sagot :
The spring constant of the spring is 11.092N/m
given that
a 180 g air-track glider is attached to a spring
the glider is pushed in 10.2 cm and released.
a student with a stopwatch finds that 12.0 oscillations take 15.0 s
The formula for calculating the period of oscillation is
T = 2[tex]\pi[/tex][tex]\sqrt{\frac{m}{k} }[/tex]
m is the mass of the spring
k is the spring constant
Making the spring constant "k" the subject of the formula will give
k = 4[tex]\pi ^{2}[/tex][tex]\frac{m}{T^{2} }[/tex]
the period of oscillation "T"
T = [tex]\frac{1}{f}[/tex]
frequency "f" is the number of oscillations completed in one second.
a student with a stopwatch finds that 12.0 oscillations take 15.0 s
number of oscillations in one sec will be 15/12 = 1.25 osc.
Period T = 1/1.25 = 0.8sec
the required spring constant;
k = 4[tex]\pi ^{2}[/tex][tex]\frac{m}{T^{2} }[/tex]
k = 4[tex](3.14)^{2}[/tex][tex]\frac{0.18}{(0.8)^{2} }[/tex]
k = 11.092N/m
Therefore, the spring constant of the spring is 11.092N/m
To learn more about spring constant :
https://brainly.com/question/14159361
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