A simple harmonic oscillator has a frequency of 2.05 Hz. What will be its amplitude of oscillation if it's started from its equillibrium position with a velocity of 1.9 m/s? Express your answer in m to three significant digits.
x = A sin 2 pi f t
x = A sin 12.9 t
v = dx/dt = 12.9 A cos 12.9 t
v = 1.9 = 12.9 A cos 0 = 12.9 A
A = 1.90/12.9 = .147 meter
To find the amplitude of oscillation, we can use the relationship between frequency and amplitude for a simple harmonic oscillator. The formula is:
ω = 2πf = √(k/m)
where ω is the angular frequency (in radians per second), f is the frequency (in Hz), k is the spring constant (in N/m), and m is the mass (in kg).
In this problem, we are given the frequency f = 2.05 Hz. We need to find the amplitude of oscillation.
Step 1: Convert the frequency from Hz to ω in radians per second.
ω = 2πf
= 2π × 2.05 Hz
≈ 12.9 rad/s (rounded to three significant digits)
Step 2: Find the mass m.
Unfortunately, the problem does not provide the mass of the oscillator. Therefore, we need more information to proceed.
To solve for the amplitude, we need to know either the mass or the spring constant.
To find the amplitude of oscillation for a simple harmonic oscillator, we need to use the equation:
ω = 2πf
where ω (omega) is the angular frequency and f is the frequency.
Given that the frequency f = 2.05 Hz, we can calculate the angular frequency as follows:
ω = 2πf = 2π × 2.05 = 12.91 rad/s (rounded to two decimal places)
Now, we can use the formula for the amplitude of oscillation:
A = v₀ / ω
where A is the amplitude, v₀ is the initial velocity, and ω is the angular frequency.
Given that the initial velocity v₀ = 1.9 m/s and ω = 12.91 rad/s, we can calculate the amplitude:
A = 1.9 / 12.91 ≈ 0.147 m (rounded to three significant digits)
Therefore, the amplitude of oscillation for the simple harmonic oscillator is approximately 0.147 m.