Question

3@5=49, 2@3=19, 9@10=271 then 6@7=?​

Answer 1

Step-by-step explanation:

A transverse standing wave is created in a string of length 150 cm and mass 1.5 kg is fixed at both ends. The tension in the string is 36 N. The string is vibrating in five loops. The maximum displacement of a particle at a distance of 35 cm from one end is given as 2mm.

Based on above information answer the following. 34.The wavelength of the wave is A) 30 cm B) 60 cm C) 15 cm D) None

35.At how many points on string the amplitude of string particle is 2 mm?

A) 3 B) 6 C) 8 D) 10

36.The maximum amplitude is

A) 6 mm B) 8 mm C) 4 mm D) None