Respuesta :
Answer:
Time period of oscillations is 0.62 s
Explanation:
Due to suspension of weight the change in the length of the spring is given as
[tex]\Delta L = L_f - L_i[/tex]
[tex]\Delta L = 21.5 - 11.9 = 9.6 cm[/tex]
now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force
[tex]mg = kx[/tex]
[tex]0.037 (9.81) = k(0.096)[/tex]
[tex]k = 3.78 N/m[/tex]
Now the period of oscillation of spring is given as
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
Now plug in all values in it
[tex]T = 2\pi \sqrt{\frac{0.037}{3.78}}[/tex]
[tex]T = 0.62 s[/tex]
The period of the oscillation for the spring with a spring constant of 3.78 N/m will be 0.7316 sec.
What is the value of the spring constant?
We know that the formula of the spring force is written as,
[tex]F = k \times \delta[/tex]
Given to us,
weight on the spring, F = 37 g = (9.81 x 0.037) = 0.36297 N
Deflection in spring, δ = (21.5 - 11.9) = 9.6 cm = 0.096 m
Substitute the value,
[tex]0.36297 = k \times 0.096\\k= 3.78\rm\ N/m[/tex]
What is the force that is been applied to the spring?
We know that the weight is been pulled down therefore, an external force is been applied to the spring to extend it further to 25.2 cm.
We know that the formula of the spring force is written as,
[tex]F = k \times \delta[/tex]
Given to us
Deflection in the spring, δ = (25.2 - 11.9) = 13.3 cm = 0.133 m
Spring constant, k = 3.78 N/m
Substitute the values,
[tex]F = 3.78 \times 0.133\\\\F = 0.5\\\\m \times g = 0.5\\\\m = 0.05125\rm\ kg[/tex]
What is the period of oscillation?
We know the formula for the period of oscillation is given as,
[tex]T = 2\pi \sqrt{\dfrac{m}{k}}[/tex]
Substitute the values,
m = 0.05125 kg
k = 3.78 N/m
[tex]T = 2\pi \sqrt{\dfrac{0.05125}{3.78}}\\T = 0.7316\rm\ sec[/tex]
Hence, the period of the oscillation for the spring with a spring constant of 3.78 N/m will be 0.7316 sec.
Learn more about Oscillation:
https://brainly.com/question/16976640
