A 277 oscillator has a speed of 122 when its displacement is 2.90 and 47.5 when its displacement is 6.80. What is the oscillator's maximum speed?
To find the oscillator's maximum speed, we need to determine the relationship between speed and displacement.
Given:
Displacement at speed 122 m/s: d₁ = 2.90 m
Displacement at speed 47.5 m/s: d₂ = 6.80 m
We can use the equation for speed in simple harmonic motion:
v = ω√(A² - x²)
where:
v represents speed,
ω represents angular frequency,
A represents the amplitude, and
x represents the displacement.
We can rearrange the equation to solve for angular frequency (ω):
ω = v / √(A² - x²)
Now, let's calculate the angular frequency (ω) when the displacement is 2.90 m:
ω₁ = 122 / √(A² - 2.90²)
And the angular frequency (ω) when the displacement is 6.80 m:
ω₂ = 47.5 / √(A² - 6.80²)
Since the angular frequency should remain constant for a given oscillator, we can equate ω₁ and ω₂:
ω₁ = ω₂
122 / √(A² - 2.90²) = 47.5 / √(A² - 6.80²)
Now, solve the above equation for A (amplitude) using algebraic manipulations:
[122 / √(A² - 2.90²)]² = [47.5 / √(A² - 6.80²)]²
(122² / (A² - 2.90²)) = (47.5² / (A² - 6.80²))
Now, cross-multiply to eliminate the square roots:
(122²)(A² - 6.80²) = (47.5²)(A² - 2.90²)
We can simplify the equation and isolate A²:
(122²)(A² - 6.80²) - (47.5²)(A² - 2.90²) = 0
Now, solve for A:
(122²)A² - (122²)(6.80²) - (47.5²)A² + (47.5²)(2.90²) = 0
Simplifying further:
(122² - 47.5²)A² = (122²)(6.80²) - (47.5²)(2.90²)
A² = [(122²)(6.80²) - (47.5²)(2.90²)] / (122² - 47.5²)
Take the square root of both sides to find the amplitude (A):
A = √{[(122²)(6.80²) - (47.5²)(2.90²)] / (122² - 47.5²)}
Now that we have the amplitude (A), we can find the maximum speed by substituting it back into the equation for speed:
v = ω√(A² - x²)
Here, we want to find the maximum speed, which occurs when the displacement (x) has its maximum value, equal to the amplitude (A).
Let's substitute A for x in the equation to get the maximum speed (vmax):
vmax = ω√(A² - A²)
Simplifying further:
vmax = ω√0
Since the square root of zero is zero, the maximum speed (vmax) of the oscillator is 0 m/s.