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An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back-andforth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?

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boffeemadrid

Answer:

2.67 m

Explanation:

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=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]

Frequency of pendulum is given by

[tex]f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}[/tex]

Given in the question

[tex]f_p=\dfrac{1}{2}f_s[/tex]

[tex]\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m[/tex]

The unstretched length of the spring is 2.67 m

samueladesida43

The unstretched length of the spring i.e. with the apple removed is 2.67m.

How to calculate length of a spring?

According to this question, the following information are given:

  • k = Spring constant = 1.5 N/m
  • g = Acceleration due to gravity = 9.81 m/s²
  • l = Unstretched length

Frequency of simple harmonic motion can be calculated using the following formula:

f(s) = ½π. √(k/m)

Frequency of pendulum is as follows:

f(p) = ½π √(g/l)

f(p) = ½ f(s)

½π √(g/l) = ½ × ½π. √(k/m)

g/l = ¼k/m

l = 4gm/k

l = 4 × 9.81 × 0.102 /1.5

l = 2.67m

Therefore, the unstretched length of the spring i.e. with the apple removed is 2.67m.

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