Question

a mass-spring system oscillates with an amplitude of 9 cm. if the force constant of the spring of 416 n/m and the mass is 0.49 kg, what is the magnitude of the maximum acceleration of the mass in m/s2?

Answer 1

The magnitude of the maximum acceleration of a mass attached to a spring is approximately 76.97 m/s^2, calculated using the formula for maximum acceleration in simple harmonic motion with the provided spring constant and mass.

Step-by-step explanation:

The magnitude of the maximum acceleration (⋅a_{max}⋅) of a mass attached to a spring during oscillations can be calculated using the formula ⋅a_{max} = ω^2 × A, where ω (omega) is the angular frequency and A is the amplitude. The angular frequency ω can be found using the formula ω = √(k/m), where k is the spring constant and m is the mass. Given the spring constant (⋅k⋅) of 416 N/m and a mass (⋅m⋅) of 0.49 kg, with an amplitude (⋅A⋅) of 9 cm (which is 0.09 m when converted), we can find ⋅a_{max}⋅.

First, calculate the angular frequency (ω) using the given values:

ω = √(k/m) = √(416 N/m / 0.49 kg) ≈ 29.15 rad/s

Then, using the angular frequency, calculate the maximum acceleration:

⋅a_{max}⋅ = ω^2 × A = (29.15 rad/s)^2 × 0.09 m ≈ 76.97 m/s^2

Therefore, the magnitude of the maximum acceleration of the mass is approximately 76.97 m/s^2.