Animated Solution for Physics - Oscillations: Two simple harmonic motion, are represented by the equations
y1=10sin(3πt+3π)
and y2=5(sin3πt+3cos3πt)
Ratio of amplitude of y1 to y2=x:1. The value of x is ............ .
Enter Numerical Value:
Visualized Solution
Amplitude of First SHM
Given first equation:
y1=10sin(3πt+3π)
Comparing with standard equation y=Asin(ωt+ϕ):
A1=10
Expanding the Second SHM
Given second equation:
y2=5(sin3πt+3cos3πt)
Expanding the bracket:
y2=5sin3πt+53cos3πt
Phasor Components
The equation has two perpendicular components:
Sine component: a=5
Cosine component: b=53
Resultant Amplitude
Resultant amplitude A2 is the vector sum:
A2=a2+b2
A2=52+(53)2
A2=25+75=100=10
Final Ratio
Ratio of amplitudes:
A2A1=1010=1
Given ratio is x:1, so:
x=1
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The Sigma Insight: Simple Harmonic Motion (SHM)
Solution Diagram
The Magic of Superposition in SHM
Simple Harmonic Motion (SHM) is one of the most elegant concepts in physics. Often, we encounter motions that are not just a simple sine wave, but a combination of sine and cosine waves. This problem is a beautiful demonstration of how we can decode such combinations.
Analyzing the First Equation
Let's start with the first equation given to us:
y1=10sin(3πt+3π)
This equation is already in the standard form of an SHM, which is y=Asin(ωt+ϕ). By directly comparing the two, we can immediately see that the amplitude of the first motion is:
A1=10
Decoding the Second Equation
The second equation looks a bit more complex because it mixes sine and cosine terms:
y2=5(sin3πt+3cos3πt)
To make sense of this, let's distribute the 5 inside the bracket:
y2=5sin3πt+53cos3πt
How do we find the amplitude of a motion that is a sum of a sine and a cosine? The secret lies in phasors. Because sine and cosine functions have a phase difference of exactly 90∘ (or π/2 radians), we can treat their coefficients as perpendicular vectors.
The Phasor Approach
Imagine a coordinate system where the horizontal axis represents the sine component and the vertical axis represents the cosine component.
Our horizontal vector has a length of a=5, and our vertical vector has a length of b=53. The resultant amplitude A2 is simply the magnitude of the vector sum of these two components. Using the Pythagorean theorem:
A2=a2+b2
Substituting our values:
A2=52+(53)2
A2=25+75
A2=100=10
The Final Conclusion
Fascinatingly, both motions have the exact same amplitude! The ratio of their amplitudes is:
A2A1=1010=1
The problem states that this ratio is x:1. Therefore, comparing the two, we find our final answer:
x=1
This problem teaches us a valuable lesson: a combination of sine and cosine waves of the same frequency is just another simple harmonic motion, shifted in phase, with an amplitude that can be easily found using vector addition.