Question

The displacement of an oscillating particle is given by x = asinωt + bcosωt where a, b and ω are constants. Prove that the particle performs a linear S.H.M. with amplitude A = √a² + b²

Pls help me with my phy que's ._. pls explain the ans step by step. Thanku T_T​

Answer 1

$\huge\green{\boxed{\textbf{\textsf{Question ↷}}}}$

The displacement of an oscillating particle is given by x = a sin ω t + b cos ω t where a, b and ω are constants.

Prove that the particle performs a linear S.H.M. with amplitude A = $\bf{\sqrt{a² + b²}}$

$\huge\green{\boxed{\textbf{\textsf{given↷}}}}$

Displacement an oscillating particle is given by x = a sin ω t + b cos ω t where a, b and ω are constants.

$\huge\green{\boxed{\textbf{\textsf{To prove↷}}}}$

Prove that the particle performs a linear S.H.M. with amplitude A = $\bf{\sqrt{a² + b²}}$

$\huge\green{\boxed{\textbf{\textsf{explanation↷}}}}$

The position of the particle is given as

y = A = cos ω t + B sin ω t ____( 1)

The Velocity of the particle is given by,

V = $\bf\dfrac{dy}{dt}$ = $\bf\dfrac{d}{dt}$ (A cos ω t + B sin ω t)

= - Aω sin ωt + Bω cos ωt

Acceleration of the particle is given by,

A = $\bf\dfrac{dv}{dt}$ (- Aω sin ωt + Bω² cos ωt)

= - Aω² cos ωt + Bω² cos ωt

= ω² (A cos ωt + B sin ωt)

= ω² = y

As the acceleration of the particle is directly proportional to the displacement & directed towards mean position... We can understand that tyhe motion is a simple harmonic motion.

Now,

Let amplitude A = r sin ᵠ ___(2)

B = r cos ᵠ ___(3)

By substituting A & B in equation (1), we get↷:

y = r sin ᵠ cos ωt + r cos ᵠ sin ωt

= r (cos ωt sin ᵠ + sin ωt cos ᵠ)

= r sin (ωt + ᵠ)

Squaring (2) & (3) and adding, we have,

A² + B² = r²

= r $\bf{\sqrt{A² + B²}}$

Dividing (2) & (3) we have,

$\bf\dfrac{A}{B}$ = tan ∅

= r $\bf{\sqrt{A² + B²}}$

$\huge\green{\boxed{\textbf{\textsf{Answer↷}}}}$

Amplitude = r $\bf{\sqrt{A² + B²}}$

Answer 2

QUESTION:-)

The displacement of an oscillating particle is given by x = asinωt + bcosωt where a, b and ω are constants. Prove that the particle performs a linear S.H.M. with amplitude A = √a² + b²?

ANSWER:-)

The displacement of an oscillating particle is given by x = a sin ω t + b cos ω t where a, b and ω are constants.

Prove that the particle performs a linear S.H.M. with amplitude A = a²+b²

Given↷

Displacement an oscillating particle is given by x = a sin ω t + b cos ω t where a, b and ω are constants.

To prove√

Prove that the particle performs a linear S.H.M. with amplitude A = a²+b²

Explanation√

The position of the particle is given as

y = A = cos ω t + B sin ω t ____( 1)

The Velocity of the particle is given by,

V = dt=dy

(A cos ω t + B sin ω t)

= - Aω sin ωt + Bω cos ωt

Acceleration of the particle is given by,

(- Aω sin ωt + Bω² cos ωt)

= - Aω² cos ωt + Bω² cos ωt

= ω² (A cos ωt + B sin ωt)

= ω² = y

As the acceleration of the particle is directly proportional to the displacement & directed towards mean position... We can understand that tyhe motion is a simple harmonic motion.

Now,

Let amplitude A = r sin ᵠ ___(2)

B = r cos ᵠ ___(3)

By substituting A & B in equation (1), we get↷:

y = r sin ᵠ cos ωt + r cos ᵠ sin ωt

= r (cos ωt sin ᵠ + sin ωt cos ᵠ)

= r sin (ωt + ᵠ)

Squaring (2) & (3) and adding, we have,

A² + B² = r²

= r \bf{\sqrt{A² + B²}}

A²+B²

Dividing (2) & (3) we have,

= tan ∅

= r \bf{\sqrt{A² + B²}}

A²+B²

Answer√

Amplitude = r \bf{\sqrt{A² + B²}}

A²+B²