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A wave is not the rope moving along/Waves · Introduction to Waves
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Time Calculation from Position vs Inverse Velocity Graph/JEE Advanced Benchmark
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Graph of position x vs inverse of velocity (v1) for a particle moving on a straight line is as shown. Find the time taken (in seconds) by the particle to move from x=3m
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to
x=15m
.
Recall the relation v=dtdx and rearrange it to find dt. Consider what the area of a horizontal strip of thickness dx and width v1 represents on the given graph.
Write the kinematic relation v=dtdx.
Rearrange the equation to express a small time interval dt as dt=v1dx.
Identify that on a graph with x on the vertical axis and v1 on the horizontal axis, the term v1dx
Conclude that the total time is the area enclosed by the curve and the vertical x-axis.
Calculate the area of the trapezium formed between x=3m and x=15m.
From the definition of velocity, we have: v=dtdx Rearranging this equation to solve for time dt, we get: dt=v1dx To find the total time t taken to move from x=3m to x=15m, we integrate both sides: t=∫315v1dx Now, let's interpret this integral graphically.
Consider a horizontal strip on the given graph at a position x with an infinitesimally small thickness dx. The width of this strip along the horizontal axis is v1. The area of this small horizontal strip is: dA=width×thickness=(v1)dx
represents the area of a horizontal strip.
Comparing this with our equation for dt, we see that dt=dA. This means the time required for a small displacement dx is equal to the area of the corresponding horizontal strip.
Therefore, the total time t is the total area enclosed by the curve and the vertical x-axis between x=3m and x=15m.
The shape of this enclosed region is a trapezium. The parallel sides are horizontal with lengths 2sm−1 and 4sm−1. The perpendicular distance (height) between these parallel sides is along the vertical axis, which is (15−3)m=12m.
