Page 393 - SEC_2017WorkingDocument_Neat

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λ
9
4 internodal loops = x = 3 m
2 2 2
4
∴ λ = m
3
-1
f = v/ λ For v = 340 ms f = 255
Hz

 The student KNOWS that there is a
Node at the wall but thinks that there is also
a Node at the speaker. [as with the vibrator
for standing waves in a string].

6
λ
5 internodal loops = 5 x = 3 m ∴ λ = m
2 5
-1
f = v/ λ For v = 340 ms f = 283 Hz
 Based on the wording of the
question: The student assumes that the 4
points are the only nodes, so that there are
Antinodes at both the wall and speaker.

λ
4 internodal loops = 4 x = 3 m ∴ λ = 1.5
2
m
-1
f = v/ λ For v = 340 ms f = 227
Hz i.e. 3 possible answers:
f = 255 Hz, 283 Hz, 227 Hz.

6b(ii) “Continuous flow method” is Heat capacity of the apparatus does not have

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