A small source of sound S having a frequency of 500 Hz is tied to the end of a light string and whirled in a vertical circle of radius 1.6 m . The string just remains tight when the source reaches the highest point. An observer is located at a large distance vertically above the centre of the circle. Determine the frequencies heard by the observer corresponding to the sound emitted by the source when it is at the exa

Step-by-step solution

At the highest point, the string just remains tight, so the tension is zero. The centripetal force is provided entirely by gravity: m v_ top ^2 R = m g v_ top = g R Using conservation of mechanical energy between the highest point and the point at the same height as the centre: 1 2 m v^2 = 1 2 m v_ top ^2 + m g R 1 2 m v^2 = 1 2 m (g R) + m g R = 3 2 m g R v = 3 g R Substituting g = 10 m s ⁻² and R = 1.6 m : v = 3 10 1.6 = 48 6.93 m s ⁻¹ When the source is at the same height as the centre, its velocity is directed