Question

The period of oscillation of a spring-and-mass system is \(0.50 \mathrm{s}\) and the amplitude is \(5.0 \mathrm{cm} .\) What is the magnitude of the acceleration at the point of maximum extension of the spring?

Short answer

Answer: The magnitude of the acceleration at the point of maximum extension of the spring is 25.13 m/s².

Step-by-step solution

Write down the known values

First, let's list down the given values: - Period of oscillation (T) = \(0.50s\) - Amplitude (A) = \(5.0cm\)

Calculate the angular frequency (ω)

We can find the angular frequency (ω) from the period using the relation: \(ω = \dfrac{2 \pi}{T}\) Now, plugging in the value of T: \(ω = \dfrac{2 \pi}{0.50s} = 4 \pi\:s^{-1}\)

Find the maximum displacement (X_max)

The maximum displacement of the mass occurs at the point of maximum extension, which is also equal to the amplitude A: \(X_{max} = 5.0 cm = 0.05 m\)

Calculate the maximum acceleration (a_max)

At the point of maximum extension, the mass experiences maximum acceleration. We can find the maximum acceleration using the general formula for the acceleration in Simple Harmonic Motion: \(a_{max} = -ω^2 X_{max}\) Now plug in the values of \(ω\) and \(X_{max}\): \(a_{max} = -(4\pi)^2 \times 0.05 = -25.13 m/s^2\) The negative sign indicates that the acceleration is opposite to the direction of the maximum displacement.

Report the magnitude of the acceleration

Since we only need to find the magnitude of the acceleration, we can ignore the negative sign: Magnitude of the acceleration at the point of maximum extension of the spring = \(25.13 \:m/s^2\)