Question:
The displacement of a particle executing SHM is \(x=3\sin2t+4\cos2t\). The amplitude of particle is
The displacement of a particle executing SHM is \(x=3\sin2t+4\cos2t\). The amplitude of particle is
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For \(x=a\sin\omega t+b\cos\omega t\), amplitude is \(\sqrt{a^2+b^2}\).
Updated On: May 7, 2026
- \(7\)
- \(3\)
- \(4\)
- \(5\)
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The Correct Option is D
Solution and Explanation
We are given:
\[
x=3\sin2t+4\cos2t.
\]
This is of the form:
\[
x=a\sin\omega t+b\cos\omega t.
\]
For such an SHM equation, amplitude is:
\[
A=\sqrt{a^2+b^2}.
\]
Here:
\[
a=3,\qquad b=4.
\]
Therefore:
\[
A=\sqrt{3^2+4^2}.
\]
\[
A=\sqrt{9+16}.
\]
\[
A=\sqrt{25}.
\]
\[
A=5.
\]
Hence, the amplitude of the particle is:
\[
5.
\]
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