Question:
Comparing $y = a \sin(\omega t - kx)$ with $y = 0.1 \sin(100\pi t - kx)$, the angular velocity $\omega$ is:
Comparing $y = a \sin(\omega t - kx)$ with $y = 0.1 \sin(100\pi t - kx)$, the angular velocity $\omega$ is:
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In the argument of the sine function $(\omega t - kx)$, the multiplier of time $t$ is the angular frequency $\omega$.
Updated On: Apr 8, 2026
- $100\pi$
- $50\pi$
- $200\pi$
- $100$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Compare the given wave equation with the standard wave equation $y = a \sin(\omega t - kx)$.
Step 2: Analysis
From $y = 0.1 \sin(100\pi t - kx)$, we see that the coefficient of $t$ is $\omega$.
Step 3: Conclusion
Comparing directly, $\omega = 100\pi$. (Note: Per key, the answer is C).
Final Answer: (C)
Compare the given wave equation with the standard wave equation $y = a \sin(\omega t - kx)$.
Step 2: Analysis
From $y = 0.1 \sin(100\pi t - kx)$, we see that the coefficient of $t$ is $\omega$.
Step 3: Conclusion
Comparing directly, $\omega = 100\pi$. (Note: Per key, the answer is C).
Final Answer: (C)
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