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ALL CIE PHYSICS DOUBTS HERE !!!

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Deadly_king:

--- Quote from: $!$RatJumper$!$ on November 09, 2010, 11:52:39 pm ---and could you also help me with s10 P22 Q4c

--- End quote ---

Jun 10 p22

4.(c) Sound with lowest frequency is produced => An antinode is found at the end of the tube and it is the only one.

Hence there should be a node at the piston. Distance between a node and the consecutive antinode is given to be lamda/4

Lambda/4 = 45 cm -----> lambda is given to be 180cm = 1.8m

V = f x lambda

Therefore f = v/lambda = 330/1.8 = 180 Hz

Frequency is given to the nearest tenth since v = 330 and lambda=1.8 (nearest 10)

Hope it helps :)

Arthur Bon Zavi:
+rep Deadly_king.

Deadly_king:

--- Quote from: Ancestor on November 10, 2010, 01:13:29 pm ---+rep Deadly_king.

--- End quote ---

Hehe...........thank you Mr the Ancestor ;D

Dania:
Refer to the attachments
Look at the paper, for the full question. [Question no. 4]
Thanks in advance.

$!$RatJumper$!$:

--- Quote from: Deadly_king on November 10, 2010, 12:34:13 pm ---Jun 10 p22

4.(c) Sound with lowest frequency is produced => An antinode is found at the end of the tube and it is the only one.

Hence there should be a node at the piston. Distance between a node and the consecutive antinode is given to be lamda/4

Lambda/4 = 45 cm -----> lambda is given to be 180cm = 1.8m

V = f x lambda

Therefore f = v/lambda = 330/1.8 = 180 Hz

Frequency is given to the nearest tenth since v = 330 and lambda=1.8 (nearest 10)

Hope it helps :)

--- End quote ---

Thank you DK :)

Another one, please explain
S09 P21 Q5b
W09 P22 Q5b

Thank you :)

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