A spring of negligible mass stretches 2.14 cm from its relaxed length when a force of 8.40 N is applied. A 1.300 kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x=5.00 cm and released from rest at t=0.
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(a) What is the force constant of the spring? (b1) What is the angular frequency, (b1) What is the frequency,
k=F/Δx=8.4/0.0214 = 393 N/m
ω= sqrt(393/1.3)=302.3 rad/s
ω=2πf
f= ω/2π=302.3/2•3.14 =48.14 Hz
(a) To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its relaxed position.
Hooke's Law equation is given by F = kx, where F is the force exerted by the spring, k is the force constant, and x is the displacement from the relaxed position.
In this case, we are given the displacement x = 2.14 cm = 0.0214 m, and the force F = 8.40 N.
Substituting these values into the equation, we can solve for the force constant k:
8.40 N = k * 0.0214 m
Dividing both sides by 0.0214 m, we get:
k = 8.40 N / 0.0214 m
Calculating this, we find:
k ≈ 392.52 N/m
Therefore, the force constant of the spring is approximately 392.52 N/m.
(b1) To find the angular frequency, we can use the equation:
Angular frequency (ω) = sqrt(k/m)
where k is the force constant of the spring and m is the mass of the particle attached to the spring.
Given k = 392.52 N/m and the mass m = 1.300 kg, we can calculate the angular frequency as follows:
ω = sqrt(392.52 N/m / 1.300 kg)
Calculating this, we find:
ω ≈ 18.64 rad/s
Therefore, the angular frequency is approximately 18.64 rad/s.
(b2) To find the frequency (f), we can convert the angular frequency (ω) to frequency (f) using the equation:
Frequency (f) = ω / (2π)
Substituting the angular frequency ω = 18.64 rad/s into the equation, we get:
f = 18.64 rad/s / (2π)
Calculating this, we find:
f ≈ 2.97 Hz
Therefore, the frequency of the oscillation is approximately 2.97 Hz.