Answer:
Some of the frequency that cannot be produced by the string includes 400Hz, 500Hz 650Hz etc...
Explanation:
Harmonics in strings are defined as the integral multiples of its fundamental frequency. This multiples are in arithmetic progression.
For example if Fo is the fundamental frequency of the string, the harmonics will be 2fo, 3fo, 4fo, 5fo... etc
If the string produces a fundamental frequency of 150Hz, some of the harmonics produced by the string will be 300Hz, 450Hz, 600Hz, 750Hz... etc
Some of the harmonics that cannot be produced include 400Hz, 500Hz 650Hz etc...
Answer:
Explanation:
In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.
To find the fundamental mode you use:
n: mode of the sound
vs: sound speed
L: length of the column of air in the tube.
A) The fundamental mode id obtained for n=1:
B) For the 3rd harmonic you have:
C) For the 2nd harmonic:
Answer:
122.735 behind converging lens ; 2.16
Explanation:
Given tgat:
Object distance, u = 29 cm
Image distance, v =
Focal length, f = - 19 (diverging lens)
Mirror formula :
1/u + 1/v = 1/f
1/29 + 1/v = - 1/19
1/v = - 1/19 - 1/29
1/v = −0.087114
v = −11.47916
v = -11.48
Second lens
Object distance :
u = 11.48 + 11 = 22.48 cm
1/v = 1/19 - 1/22.48
1/v = 0.0081475
v = 1 / 0.0081475
v = 122.735 cm
122.735 behind second lens
Magnification, m
m = m1 * m2
m = - v / u
Lens1 :
m1 = -11.48 / 29 = - 0.3958620
m2 = - 122.735 / 22.48 = - 5.4597419
Hence,
- 0.3958620 * - 5.4597419 = 2.16
