Question

A 210 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz. At t = 0 s, the mass is at x = 5.60 cm and has vx = -31.0 cm/s. What is the amplitude of the oscillation?

1) 2.80 cm
2) 5.60 cm
3) 11.20 cm
4) 22.40 cm

Answer 1

The amplitude of the oscillation is found by equating the kinetic energy at the given point to the potential energy at the amplitude. After solving the equations, the amplitude is determined to be 5.60 cm, which matches the provided option 2).

Step-by-step explanation:

To find the amplitude of the oscillation, we need to analyze the simple harmonic motion described in the question. At any point in time, the total energy of the system will be the sum of its potential and kinetic energies, which remains constant (ignoring any damping). Using the information given, we know the mass m is 210 g (or 0.210 kg), the frequency f is 3.40 Hz, and the initial velocity vx is -31.0 cm/s (or -0.310 m/s) at an initial position x of 5.60 cm (0.056 m).

The potential energy (PE) of the spring at any point is PE = 1/2 k x2, and the kinetic energy (KE) is KE = 1/2 m v2. At the amplitude of the motion, all the energy is potential, hence:

PE = KE
1/2 k A2 = 1/2 m v2

And to find k, we use the relationship between frequency and the spring constant with mass:

f = 1/2π √(k/m)

Solving this will give us k, and we can solve for A, the amplitude.

After finding k and substituting in the potential energy equation and equating it to the kinetic energy:

A2 = m v2/k

We can solve for A, and the closest value provided in the options is 5.60 cm, which is option 2).