An object with mass 2.0kg is attached to a spring with spring stiffness constant 240N/m and is executing simple harmonic motion. When the object is 2.2×10−2m from its equilibrium position, it is moving with a speed of 0.65m/s . Calculate the amplitude of the motion.
x=2.2•10⁻² m
v=0.65 m/s
m=2 kg
k =240 N/m =>
ω=sqrt(k/m) = sqrt(240/2) =11 rad/s
x=Asin ωt
v= ω Acos ωt
tan ωt = xω /v=2.2•10⁻²•11/0.65 =0.372
ωt =20.4°
sin ωt=0.349
x=Asin ωt
A=x/sin ωt =2.2•10⁻²/0.349=6.3•10⁻² m
To calculate the amplitude of the motion, we need to use the formulas of simple harmonic motion and apply the given information.
The formula for the speed of an object undergoing simple harmonic motion is given by:
v = ω √(A² - x²)
Where:
v = speed of the object (0.65 m/s in this case)
ω = angular frequency of the motion (which can be calculated using the formula ω = √(k/m), where k is the spring constant and m is the mass of the object)
A = amplitude of the motion (the value we are trying to find)
x = displacement of the object from its equilibrium position (2.2×10^(-2) m in this case)
First, let's calculate the angular frequency using the given spring constant and mass:
ω = √(k/m)
ω = √(240 N/m ÷ 2.0 kg)
ω = √(120 N/kg)
ω ≈ 10.95 rad/s (rounded to two decimal places)
Now, we can rearrange the formula for the speed of the object to solve for the amplitude:
v = ω √(A² - x²)
Rearranging the formula gives:
(A² - x²) = (v/ω)²
Substituting the given values, we have:
(A² - (2.2×10^(-2))²) = (0.65/10.95)²
(A² - 4.84×10^(-4)) = 3.411×10^(-3)
A² = 4.84×10^(-4) + 3.411×10^(-3)
A² ≈ 3.895×10^(-3)
Taking the square root of both sides:
A ≈ √(3.895×10^(-3))
A ≈ 0.0624 m (rounded to four decimal places)
Therefore, the amplitude of the motion is approximately 0.0624 meters.