11 grade physics others

A point mass oscillates along the x-axis according to the law x = x₀cos(ωt - π/4). If the acceleration of the particle is written as a = Acos(ωt + δ), then
A. A = x₀, δ = π/4
B. A = x₀ω², δ = π/4
C. A = x₀ω², δ = -π/4
D. A = x₀ω², δ = 3π/4

Aniket Singh

10 Months agoGrade

1 Answer

Askiitians Tutor Team

Approved Tutor Answer10 Months ago

The motion of a point mass oscillating along the x-axis can be described by the equation x = x₀cos(ωt - π/4). To find the acceleration, we need to differentiate the position function twice with respect to time.

Finding the Acceleration

The first derivative of the position function gives us the velocity:

v = dx/dt = -x₀ωsin(ωt - π/4)

Next, we differentiate the velocity to find the acceleration:

a = dv/dt = -x₀ω²cos(ωt - π/4)

Rewriting the Acceleration

The acceleration can be expressed in the form a = Acos(ωt + δ). To match this with our derived expression:

The amplitude A is x₀ω².
The phase shift δ is -π/4.

Conclusion

Thus, the correct answer is C. A = x₀ω², δ = -π/4.