HSC Science (Electronics) 12th Board ExamMaharashtra State Board
Share

A Set of 12 Tuning Forks is Arranged in Order of Increasing Frequencies. Each Fork Produces ‘Y’ Beats per Second with the Previous One. the Last is an Octave of the First. the Fifth Fork Has a Frequency of 90 Hz. Find ‘Y’ and Frequency of the First and the Last Tuning Forks. - HSC Science (Electronics) 12th Board Exam - Physics

Question

A set of 12 tuning forks is arranged in order of increasing frequencies. Each fork produces
‘Y’ beats per second with the previous one. The last is an octave of the first. The fifth fork
has a frequency of 90 Hz. Find ‘Y’ and frequency of the first and the last tuning forks.

Solution

Solution:
Given: N = 12, nL = 2 nF , n5 = 90 Hz

To find: Number of beats (x),

Frequency of last fork (nL)
Formula: nL = nF + (N - 1) x

nL = nF + (5- 1) x
nF + 4x = 90 ….(1)
nL = nF + (12 - 1) x
nL = nF + 11x ….(2)
nL= 2nF
nF= 11x ….From (2)
Substituting in equation (1),
15x = 90 or x = 6 beat/s
nF = 11 x 6 = 66 Hz
nL = 2 x 66 = 132 Hz
The frequency of the first and last tuning fork is 66 Hz and 132 Hz respectively.

Is there an error in this question or solution?

APPEARS IN

2016-2017 (July) (with solutions)
Question 2.3 | 3.00 marks

Video TutorialsVIEW ALL [1]

Solution A Set of 12 Tuning Forks is Arranged in Order of Increasing Frequencies. Each Fork Produces ‘Y’ Beats per Second with the Previous One. the Last is an Octave of the First. the Fifth Fork Has a Frequency of 90 Hz. Find ‘Y’ and Frequency of the First and the Last Tuning Forks. Concept: Formation of Beats.
S