# loudspeaker experiment MCQ

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#1
Hi,

I need some help on the following question.

The answer is G, but I have no clue on how to get to it. Is it to do with stationary waves?

0
4 weeks ago
#2
(Original post by Navboi)
Hi,

I need some help on the following question.

The answer is G, but I have no clue on how to get to it. Is it to do with stationary waves?

Think about the delay time for both echoes to be heard by the student for a single emitted click.

There will be a time delay between the echoes.

In order for the perceived echoes to coincide, repeated clicks will need to correspond to the echo delay difference. Voila.
Last edited by uberteknik; 4 weeks ago
0
4 weeks ago
#3
This is an interesting question. The answer has nothing to do with stationary waves, since there is no interference pattern. You could start by considering the travel and return times of the waves from each of the two reflecting walls. Remember that the waves are emitted in all directions. You can then turn each time in to a frequency, which represents the minimum 'fundamental' frequency at which clicks will received at the same time as they are emitted (it will be different for each). Try to see if you can then take the final step to getting the frequency at which clicks from both walls are received simultaneously.
1
#4
(Original post by uberteknik)
Think about the delay time for both echoes to be heard by the student for a single emitted click.

There will be a time delay between the echoes.

In order for the perceived echoes to coincide, repeated clicks will need to correspond to the echo delay difference. Voila.
Right, so I worked out the time difference to be 0.1 seconds. But I don't really understand how this could be changed to the frequency of the clicks. (I know its probably something like 1/0.1 = 10hz but Idk what steps you'd follow to come to that conclusion)
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4 weeks ago
#5
(Original post by Navboi)
Right, so I worked out the time difference to be 0.1 seconds. But I don't really understand how this could be changed to the frequency of the clicks. (I know its probably something like 1/0.1 = 10hz but Idk what steps you'd follow to come to that conclusion)
The time difference is not 0.1 seconds. Remember the waves have to travel to the wall and back again before they can be heard.

It might be easier to consider the travel-and-return time for each wave as a frequency (without taking the difference) and then see if there is a lowest common multiple of them.
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