Question

a body of mass m is executing simple harmonic motion with an amplitude of 8.0 cm and a maximum acceleration of 100 cm/s2. when the displacement of this body from the equilibrium position is 6.0 cm, the magnitude of the acceleration is approximately

Answer 1

Answer: Magnitude of acceleration: 937.5 cm/s^2

Step-by-step explanation:

To find the magnitude of acceleration at a given displacement in simple harmonic motion, we can use the equation:

a = -ω²x

Where:

a is the acceleration,

ω (omega) is the angular frequency, and

x is the displacement from the equilibrium position.

In this case, we are given the amplitude (A) and the maximum acceleration (a_max). The maximum acceleration is equal to ω²A, so we can rearrange the equation to find ω:

ω = √(a_max / A)

Substituting the given values:

a_max = 100 cm/s²

A = 8.0 cm

ω = √(100 cm/s² / 8.0 cm) = √12.5 rad/s

Now we can find the magnitude of acceleration at a displacement of 6.0 cm:

x = 6.0 cm

a = -ω²x = -(12.5 rad/s)² * (6.0 cm) ≈ -937.5 cm/s²

Therefore, the magnitude of the acceleration at a displacement of 6.0 cm is approximately 937.5 cm/s².