The period of oscillation of an object in an ideal spring-and-mass system is \(0.50 \mathrm{s}\) and the amplitude is \(5.0 \mathrm{cm} .\) What is the speed at the equilibrium point?

We are given: - Period of oscillation, \(T = 0.50 \;\text{s}\) - Amplitude of oscillation, \(A = 5.0 \;\text{cm} = 0.050 \;\text{m}\) (converting to meters)

We can find the angular frequency (\(\omega\)) using the period \(T\) with the formula: \(\omega = \frac{2\pi}{T}\) Substituting the given values, we get: \(\omega = \frac{2\pi}{0.50 \;\text{s}}\) Now, calculate \(\omega\): \(\omega = 12.57 \;\text{rad/s}\)

At the equilibrium point, the displacement of the object is zero (\(x = 0\)), which means that the speed will be maximum. The formula for the speed of an object in a spring-mass system is given by: \(v = A\omega\sin(\omega t + \phi)\) At the equilibrium point, the phase angle is zero (\(\phi = 0\)) and the time (\(t\)) at which the object is at the equilibrium point is given by a multiple of half of the period (\(t=n\frac{T}{2} ,\) n is an integer). Since \(\sin(\omega t + \phi)\) will be maximum (equal to 1) at the equilibrium point, the speed of the object will be: \(v = A\omega\) Substitute the given values: \(v = (0.050 \;\text{m})(12.57 \;\text{rad/s})\) Now, calculate the speed: \(v = 0.6285 \;\text{m/s}\)

The speed of the object at the equilibrium point in this ideal spring-and-mass system is \(0.6285\;\text{m/s}\).