Question

1. a 2 kg body is attatched to a spring of negligible mass and oscillates with a period of 1.00 s. The force constant of the spring is?

2. The force constant of a massless spring is 25 N/m. A mass of .45 kg is oscillating in simple harmonic motion at the end of the spring with an amplitude of .32 m. The maximum speed of the mass is?

please explain

Answer 1

1. To find the force constant (k) of the spring, we can use the formula for the period of oscillation of a mass-spring system:

T = 2 * pi * sqrt(m/k)

where T is the period of oscillation, m is the mass, and k is the spring constant (force constant). We are given T and m, and we need to find k.

1.00 s = 2 * pi * sqrt(2 kg / k)

To solve for k, first divide both sides of the equation by 2 * pi:

(1.00 s) / (2 * pi) = sqrt(2 kg / k)

Now, square both sides to get rid of the square root:

[(1.00 s) / (2 * pi)]^2 = 2 kg / k

Now, solve for k:

k = 2 kg / [(1.00 s) / (2 * pi)]^2

k ≈ 39.48 N/m

So, the force constant of the spring is approximately 39.48 N/m.

2. To find the maximum speed (v_max) of the mass, we can use the formula for the maximum speed in simple harmonic motion:

v_max = A * ω

where A is the amplitude, and ω is the angular frequency. We are given the amplitude (0.32 m). To find the angular frequency, we can use the formula for the angular frequency in a mass-spring system:

ω = sqrt(k/m)

where k is the force constant (25 N/m), and m is the mass (0.45 kg). So,

ω = sqrt(25 N/m / 0.45 kg) ≈ 7.43 rad/s

Now, we can find the maximum speed:

v_max = 0.32 m * 7.43 rad/s ≈ 2.38 m/s

So, the maximum speed of the mass is approximately 2.38 m/s.