In simple harmonic motion, the maximum acceleration (Amax) is given by the formula:

Amax = ω² * A

where ω is the angular frequency and A is the amplitude of the motion.

From the problem, we know that the ratio of maximum acceleration to maximum velocity (Vmax) is 10 s^-1. The maximum velocity in simple harmonic motion is given by:

Vmax = ω * A

So, we can write the given ratio as:

Amax/Vmax = ω = 10 s^-1

We also know that at t = 0, the displacement x = A * cos(Φ), where Φ is the initial phase. Substituting the given values:

5 m = A * cos(π/4)

Solving for A, we get:

A = 5 m / cos(π/4) = 5√2 m

Now, we can find the maximum acceleration using the formula Amax = ω² * A:

Amax = (10 s^-1)² * 5√2 m = 500√2 m/s²

So, the maximum acceleration is 500√2 m/s².

Question

In simple harmonic motion, the maximum acceleration (Amax) is given by the formula:

Amax = ω² * A

where ω is the angular frequency and A is the amplitude of the motion.

From the problem, we know that the ratio of maximum acceleration to maximum velocity (Vmax) is 10 s^-1. The maximum velocity in simple harmonic motion is given by:

Vmax = ω * A

So, we can write the given ratio as:

Amax/Vmax = ω = 10 s^-1

We also know that at t = 0, the displacement x = A * cos(Φ), where Φ is the initial phase. Substituting the given values:

5 m = A * cos(π/4)

Solving for A, we get:

A = 5 m / cos(π/4) = 5√2 m

Now, we can find the maximum acceleration using the formula Amax = ω² * A:

Amax = (10 s^-1)² * 5√2 m = 500√2 m/s²

So, the maximum acceleration is 500√2 m/s².

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