Question

A mass of 4.0 kg is placed on the end of a spring. The graph provided indicates the mass' potential energy as a function of distance from the equilibrium position. The total mechanical energy of the system is 300 J. What is the value of the spring constant when the mass is -2.0 m from the equilibrium position? Potential Energy vs. Distance 400 2 30 100- -2.010 0 1.0 2.0 Distance (m) A) 25 N/m B) 35 N/m C) 75 N/m D) 150 N/m 23. A string is firmly attached at both ends. When a frequency of 60 Hz is applied, the string vibrates in the standing wave pattern shown. Assume the tension in the string and its mass per unit length do not change. Which of the following frequencies could NOT also produce a standing wave pattern in the string? A) 30 Hz B) 40 Hz C) 80 Hz D) 180?

Answer 1

Total energy = 300 J so, maximum potential energy = 300 J so, 1/2 kx^2 max = 300 1/2 k times (-2)^2 = 300 1/2 k times (-2)^2 = 300 as x_max = 2 m 2 k = 300 k = 150 N/m(option - D) we know the general formula for me frequency of a string attached at both ends V_n = n/2l squareroot T/mu T = tension mu = mass per unit length As here are three loops, so, n = 3 so, V_n = 3 middot V_0 V_0 = V_n/3 = 60/3 = 20 Hz V_0 = 1/2l squareroot T/mu = fundamental frequency so to create a standing wave the vibrating frequency must have frequency which is an integral multiple of V_0(= 20 Hz) so, here 30 Hz is not a integral multiple of 20 Hz therefore option - A is correct.