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Physics · JEE | OSCILLATIONS

हिन्दी

If a periodic function is expressed as \( f(t) = A\sin(\omega t) + B\cos(\omega t) \), it can be rewritten in the form \( f(t) = D\sin(\omega t + \phi) \). What is the value of amplitude \( D \)?

  1. A
    A + B
  2. B
    \( \sqrt{A^2 - B^2} \)
  3. C
    \( \frac{A}{B} \)
  4. D
    \( \sqrt{A^2 + B^2} \)
View Answer & Solution

Correct answer · Option D

\( \sqrt{A^2 + B^2} \)

Explanation

By taking \( A = D\cos\phi \) and \( B = D\sin\phi \), the resultant amplitude is \( D = \sqrt{A^2 + B^2} \).
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