Question

the displacement of the particle executing simple harmonic motion is given by Y = Ao + A Sin Omega T + B cos Omega T then the amplitude of the oscillation is given by​

Answer 1

Answer:

The amplitude of its oscillation is given by

\bold{\sqrt{A^{2}+B^{2}}}

A

2

+B

2

Step by step explanation:

"Simple harmonic motion in which restoring force is directly proportional to the displacement of the particle from the mean or equilibrium position."

From the given,

The displacement of a particle executing simple harmonic motion is given by

\bold{y=A_{o}+A\,sin\omega t+B\,cos\omega t}y=A

o

+Asinωt+Bcosωt

y^{I}=y-A_{o}y

I

=y−A

o

\Rightarrow A\,sin\,\omega t+B\,cos\,\omega t⇒Asinωt+Bcosωt

From the diagram, (in attachment)

R=\sqrt{A^{2}+B^{2}+2AB\,cos\,90^{o}}R=

A

2

+B

2

+2ABcos90

o

\Rightarrow \sqrt{A^{2}+B^{2}}⇒

A

2

+B

2

The amplitude of its oscillation is given by \bold{\sqrt{A^{2}+B^{2}}}

A

2

+B

2

.