Question

If a mass is vibrating at a frequency of 0.9 Hz and undergoes a speed of 0.05 m/s as it passes through the center of oscillation. What is its amplitude? What is the oscillation period?

Answer 1

To determine the amplitude and oscillation period, we can use the given frequency and speed of the vibrating mass.

1. Amplitude: The speed of the mass as it passes through the center of oscillation is equal to the maximum velocity of the vibrating mass. In simple harmonic motion, the maximum velocity is related to the amplitude (A) and angular frequency (ω) by the equation: v_max = Aω.

In this case, the frequency (f) is given as 0.9 Hz. The angular frequency (ω) can be calculated using the equation: ω = 2πf.

So, ω = 2π × 0.9 = 5.67 rad/s.

Now, using the given speed of 0.05 m/s, we can solve for the amplitude: 0.05 m/s = A × 5.67 rad/s. A = 0.05 m/s / 5.67 rad/s ≈ 0.0088 m (or 8.8 mm).

Therefore, the amplitude of the oscillation is approximately 0.0088 meters or 8.8 millimeters.

2. Oscillation Period: The period (T) of oscillation is the time taken for one complete cycle of the motion. It is the inverse of the frequency: T = 1/f.

In this case, the frequency is given as 0.9 Hz. T = 1 / 0.9 = 1.11 seconds (approximately).

Therefore, the oscillation period is approximately 1.11 seconds.