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Find the average speed of a particle whose velocity is given by V= v0sin ​​omega t, where T= 2pi/omega is the time of complete cycle.
Solution Vav=(1/T)×∫02pi/omega v0sin omega t dt
=(1/2 pi/ omega) ×( ∫0pi/omega v0sin omega t dt - ∫pi/omega2pi/omega v0 sin omega t dt) (sin 0=negative for 0= pi to 2pi)
In the solution to the above problem why the limits of integration are taken as first 0 to pi/omega minus integration with limits pi/omega to 2pi/ omega?
Why can't we just directly integrate from 0 to 2pi/omega?

Pratyay Talukdar , 9 Years ago
Grade 11
anser 1 Answers
Askiitians Tutor Team

Last Activity: 5 Months ago

To find the average speed of a particle whose velocity is defined by the equation \( V = v_0 \sin(\omega t) \), we need to consider the nature of the sine function and how it behaves over one complete cycle. The average speed is calculated using the formula:

Understanding Average Speed Calculation

The average speed \( V_{av} \) over a time period \( T \) is given by:

V_{av} = \frac{1}{T} \int_0^T V \, dt

In this case, \( T = \frac{2\pi}{\omega} \), which represents the time for one complete cycle of the sine wave. Plugging this into our formula gives:

V_{av} = \frac{1}{\frac{2\pi}{\omega}} \int_0^{\frac{2\pi}{\omega}} v_0 \sin(\omega t) \, dt

Why Split the Integral?

Now, let's address your question about why we split the integral into two parts: from \( 0 \) to \( \frac{\pi}{\omega} \) and from \( \frac{\pi}{\omega} \) to \( \frac{2\pi}{\omega} \).

  • Behavior of the Sine Function: The sine function oscillates between positive and negative values. Specifically, from \( 0 \) to \( \pi \), \( \sin(\omega t) \) is positive, while from \( \pi \) to \( 2\pi \), it becomes negative.
  • Impact on Average Speed: If we were to integrate directly from \( 0 \) to \( \frac{2\pi}{\omega} \), the negative values from the second half of the cycle would cancel out the positive values from the first half, leading to an incorrect calculation of average speed.

Calculating the Integral

To compute the average speed correctly, we can evaluate the two integrals separately:

1. First Integral:

From \( 0 \) to \( \frac{\pi}{\omega} \):

∫_0^{\frac{\pi}{\omega}} v_0 \sin(\omega t) \, dt

This integral will yield a positive value since \( \sin(\omega t) \) is positive in this range.

2. Second Integral:

From \( \frac{\pi}{\omega} \) to \( \frac{2\pi}{\omega} \):

∫_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} v_0 \sin(\omega t) \, dt

This integral will yield a negative value, reflecting the negative sine values in this range.

Final Calculation

By splitting the integral, we can accurately account for the contributions of both halves of the cycle. The average speed can then be computed as:

V_{av} = \frac{1}{\frac{2\pi}{\omega}} \left( \int_0^{\frac{\pi}{\omega}} v_0 \sin(\omega t) \, dt - \int_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} v_0 \sin(\omega t) \, dt \right)

This approach ensures that we correctly represent the particle's motion over the entire cycle, leading to an accurate calculation of average speed. By understanding the behavior of the sine function and the importance of integrating over the correct intervals, we can derive meaningful results in physics.

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