A small particle of mass 2.0 x 10^-19 kg is attached to a string of relaxed length 0.50 nm .The string is the black horizontal line attached to a an immovable wall on the left. The plot is a graph of the force needed to stretch the string by pulling to the right on it .The length of the string is x.

a) Calculate the force constant-the “spring constant” k for this string.
b) Assume that the string 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. Can some one show me how to put this in a formula and solve?

a) To calculate the force constant (spring constant) k, we can use Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement from its equilibrium position.

Hooke's Law can be expressed as: F = -kx

Where F is the force applied, x is the displacement from the equilibrium position, and k is the spring constant.

In this scenario, we have a mass attached to a string being stretched, so the force required to stretch the string is equal to the gravitational force acting on the particle.

The force due to gravity is calculated as: F_gravity = mg

Given that the mass of the particle is 2.0 x 10^-19 kg, we can calculate the force due to gravity as: F_gravity = (2.0 x 10^-19 kg)(9.8 m/s^2) = 1.96 x 10^-18 N

Since the force required to stretch the string is equal to the force due to gravity, we can equate the two expressions:

-kx = F_gravity
-k(0.5 x 10^-9 m) = 1.96 x 10^-18 N

Solving for k:
k = (1.96 x 10^-18 N) / (0.5 x 10^-9 m)
k = 3.92 x 10^-9 N/m

Therefore, the spring constant (force constant) for this string is k = 3.92 x 10^-9 N/m.

b) To calculate the maximum speed of the particle as it oscillates back and forth, we can use the equation for Simple Harmonic Motion:

v_max = Aω

Where v_max is the maximum velocity (speed), A is the amplitude (maximum displacement), and ω is the angular frequency.

The angular frequency ω can be calculated as: ω = √(k / m)

Given the spring constant k = 3.92 x 10^-9 N/m and the mass of the particle m = 2.0 x 10^-19 kg, we can calculate ω:

ω = √((3.92 x 10^-9 N/m) / (2.0 x 10^-19 kg))
ω = √(1.96 x 10^10 N/kg)
ω ≈ 4.43 x 10^5 rad/s

Now, let's assume that the string is stretched to a total length of x = 3.0 nm and released from rest. The maximum displacement (amplitude) A can be calculated as half of the total length stretched:

A = x / 2
A = (3.0 x 10^-9 m) / 2
A = 1.5 x 10^-9 m

Plugging the values of A and ω into the equation for maximum velocity:

v_max = (1.5 x 10^-9 m)(4.43 x 10^5 rad/s)
v_max ≈ 6.65 m/s

Therefore, the maximum speed of the particle as it oscillates back and forth is approximately 6.65 m/s.

a) To calculate the force constant or the spring constant (k) for this string, we can use Hooke's Law.

Hooke's Law states that the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position, and is given by the equation:

F = -kx

Where:
F = force (N)
k = spring constant or force constant (N/m)
x = displacement (m)

In this case, the string is stretched by pulling to the right, so the displacement is positive. The force needed to stretch the string is the force constant multiplied by the displacement (F = kx).

Given:
Mass of the particle (m) = 2.0 x 10^-19 kg
Relaxed length of the string (x0) = 0.50 nm

To calculate the force constant (k), we can rearrange Hooke's Law equation:

k = F/x

We know the force (F) needed to stretch the string by pulling to the right is given by:

F = mg

Where g is the acceleration due to gravity.

Substituting the values:

F = (2.0 x 10^-19 kg)(9.8 m/s^2)

Simplifying, we get:

F = 1.96 x 10^-18 N

Now, substituting the values into the equation for k:

k = (1.96 x 10^-18 N)/(0.50 x 10^-9 m)

Simplifying, we get:

k = 3.92 x 10^10 N/m

So, the force constant or spring constant (k) for this string is 3.92 x 10^10 N/m.

b) To compute the maximum speed of the particle as it oscillates back and forth, we can use the equation for simple harmonic motion:

v = √(2πk/m) * A

Where:
v = maximum speed (m/s)
k = spring constant (N/m)
m = mass of the particle (kg)
A = amplitude (maximum displacement) (m)

Given:
k = 3.92 x 10^10 N/m
m = 2.0 x 10^-19 kg
Total length x = 3.0 nm = 3.0 x 10^-9 m

First, let's calculate the amplitude (A):

Since the total length (x) is the sum of the relaxed length (x0) and the amplitude (A), we can write:

x = x0 + A

3.0 x 10^-9 m = 0.50 x 10^-9 m + A

A = (3.0 - 0.50) x 10^-9 m

A = 2.5 x 10^-9 m

Now, substituting the values into the equation for maximum speed (v):

v = √(2π(3.92 x 10^10 N/m)/(2.0 x 10^-19 kg)) * (2.5 x 10^-9 m)

Simplifying, we get:

v = √(2π(3.92 x 10^10) * (2.5 x 10^-9))

v = √(3.92 x 2.5 x 2π) x 10^10-9 N/kg m

v = √(24.57π) x 10^1 N/kg

v ≈ 15.67 m/s

So, the maximum speed of the particle as it oscillates back and forth is approximately 15.67 m/s.