for a particle executing simple harmonic motion the displacement is 8 cm at the instant the velocity is 6cm\sec and the displacement is 6 cm at the instant the velocity is 8 cm\sec calculate the amplitude?
To calculate the amplitude of a particle executing simple harmonic motion, we can use the relationship between displacement and velocity at any given instant.
In simple harmonic motion, the displacement (x) and velocity (v) of the particle are related by the equation:
v = ω√(A² - x²)
Where:
v = velocity of the particle
A = amplitude
x = displacement of the particle
ω = angular frequency of the motion (ω = 2πf, where f is the frequency)
Let's solve the equations given in the question using this equation to find the amplitude (A).
Given:
First instant - displacement (x) = 8 cm, velocity (v) = 6 cm/sec
Second instant - displacement (x) = 6 cm, velocity (v) = 8 cm/sec
Using the equation v = ω√(A² - x²), let's solve for ω first.
First instant:
6 = ω√(A² - 8²)
Square both sides of the equation:
36 = ω²(A² - 64)
Second instant:
8 = ω√(A² - 6²)
Square both sides of the equation:
64 = ω²(A² - 36)
We have two equations with two unknowns (A and ω). We can solve these simultaneously.
Rewriting the two equations:
36 = ω²(A² - 64) ---> Equation 1
64 = ω²(A² - 36) ---> Equation 2
Divide Equation 1 by Equation 2:
36/64 = (A² - 64)/(A² - 36)
Simplifying further:
9/16 = (A² - 64)/(A² - 36)
Cross-multiply:
9(A² - 36) = 16(A² - 64)
Expanding and rearranging the equation:
9A² - 324 = 16A² - 1024
Combine like terms:
7A² = 700
Divide by 7:
A² = 100
Take the square root of both sides:
A = √100
A = 10 cm
Therefore, the amplitude of the particle executing simple harmonic motion is 10 cm.