5: Sam, a butcher, puts 3.0 kg of chopped beef on the 1.0-kg pan of his scale, which has a spring whose force constant is 400.N/m What is the period of vibration of the pan as it comes to rest? b) If Sam adds more beef to the scale, what will this do to the period of vibration?

To solve this problem, we need to find the period of vibration of the pan as it comes to rest, and then determine how adding more beef to the scale will affect the period of vibration.<br /><br />Given information:<br />- Mass of the pan: 1.0 kg<br />- Spring constant: 400 N/m<br />- Mass of chopped beef: 3.0 kg<br /><br />a) To find the period of vibration of the pan as it comes to rest, we can use the formula for the period of a simple harmonic oscillator:<br /><br />T = 2π√(m/k)<br /><br />Where:<br />- T is the period of vibration<br />- m is the mass of the system (pan + beef)<br />- k is the spring constant<br /><br />Substituting the given values:<br />T = 2π√((1.0 kg + 3.0 kg) / 400 N/m)<br />T = 2π√(4.0 kg / 400 N/m)<br />T = 2π√(0.01 kg/N)<br />T = 2π√(0.01 s²)<br />T = 2π × 0.1 s<br />T = 0.628 s<br /><br />Therefore, the period of vibration of the pan as it comes to rest is 0.628 seconds.<br /><br />b) If Sam adds more beef to the scale, the period of vibration will increase.<br /><br />The reason for this is that the period of vibration is directly proportional to the square root of the mass of the system. When more beef is added to the scale, the mass of the system increases, which in turn increases the period of vibration.<br /><br />In other words, as the mass of the system increases, the period of vibration increases, and vice versa. This relationship is a fundamental property of simple harmonic oscillators, such as the pan and spring system in this problem.