Question

The diagrams given below show two sound boxes A and B with wires of same length (l) and tension (10 kgf) but different cross-sectional areas. Simultaneously, vibrating tuning forks of frequency 300 Hz are placed on the boxes A and B. The paper rider falls off in case of B but not in case of A.

1. Name and explain the phenomenon responsible for the falling off of the paper rider in B.
2. The wire A resonates with a tuning fork of frequency ‘f’. Is ‘f’ greater than, less than or equal to 300 Hz? Justify your answer.

Answer 1

1. The phenomenon occurring is resonance in stretched strings. When the frequency of the applied force of the tuning fork matches or is an integral multiple of the natural frequency of the stretched string, the body vibrates with maximum amplitude, thus, the paper rider comes off.
2. According to the law of vibrating strings, the frequency of emission is inversely proportional to the square root of the linear density of the string, i.e., the thickness of the string. Given that wire A is thicker and has a lower resonance frequency, its tuning fork frequency (f) will be lower than 300 Hz.