A small particle of mass 2.0 x 10^-19 kg is attached to a nanofiber of relaxed length 0.50 nm (Recall that 1.0 nm + 10^-9 m) . The nanofiber is the black horizontal line-its attached to a substrate (an immovable wall) on the left. The plot is a graph of the force needed to stretch the nanofiber by pulling to the right on it. Let’s just call the length of the nanofiber for simplicity.
a) calculate the force constant-the “spring constant” k for this nanofiber.
b) assume that the nanofiber is stretched to a total length of x = 3.0 nm and released from rest. Then the particle will move back and forth along the x direction. Use your k from part a to compute the maximum speed of the particle as it oscillates back and forth.
a) To calculate the force constant (spring constant) of the nanofiber, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law equation: F = -kx
Where:
F is the force applied to the spring (in newtons),
k is the force constant (in newtons per meter),
x is the displacement of the spring from its equilibrium position (in meters).
We are given:
Mass of the particle (m) = 2.0 x 10^-19 kg
Relaxed length of the nanofiber (x₀) = 0.50 nm = 0.50 x 10^-9 m
Force constant can be calculated using the following formula:
k = mω²
Where:
m is the mass of the particle (in kilograms),
ω is the angular frequency (in radians per second).
To find ω, we can use the formula for the angular frequency of a simple harmonic oscillator:
ω = √(k / m)
Rearranging the equation, we can solve for k:
k = mω² = m(2πf)²
Since we are given the length of the nanofiber and not the frequency, we need to calculate the frequency first.
The total length of the stretched nanofiber (x) is given as 3.0 nm = 3.0 x 10^-9 m.
Since the particle is attached to an immovable wall, it experiences a restoring force when displaced from its equilibrium position. Therefore, it undergoes simple harmonic motion.
The maximum displacement (A) of the particle from its equilibrium position can be calculated using the equation:
A = x - x₀
Substituting the values, we get:
A = (3.0 x 10^-9 m) - (0.50 x 10^-9 m) = 2.5 x 10^-9 m
The maximum speed (v_max) of the particle can be calculated using the formula:
v_max = Aω
Substituting the values, we get:
v_max = (2.5 x 10^-9 m)(2πf)
Now we have all the necessary information to solve the problem.