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

For the string to just remain tight at the highest point, the tension must be zero. The centripetal force is provided entirely by gravity: mv_ top ^2 R = mg v_ top = gR Using the conservation of mechanical energy, we can find the velocity of the source at the lowest point of the circular path: 1 2 mv_ bot ^2 = 1 2 mv_ top ^2 + mg(2R) v_ bot = v_ top ^2 + 4gR = gR + 4gR = 5gR Substituting the given values g = 10 m s ⁻² and R = 1.6 m : v_ bot = 5 10 1.6 = 80 8.94 m s ⁻¹ From the given figure, the source is moving in