Physics - Sound

bonkysleuth SG

A piece of thin card is held against the teeth of a cog wheel. When the wheel is turned at high speed, a note is heard. How can the pitch of the note be raised?

A By pressing card against teeth with greater force

B By turning wheel more quickly

C By using a thicker card

D By using a wheel with fewer teeth

My guess is B or D. But D seems more likely because the question already stated the wheel is turned at high speed. And with fewer teeth, the number of waves generated be second will consequently be smaller.

Next question. A man who is standing at a point X between 2 parallel walls fires a starting pistol.

He hears the 1st echo after 0.6 secs and another one after 0.8secs/

How long after firing the pistol will he hear the next echo?

A 1.2s

B 1.4s

C 1.6s

D 2.4s

eagle SG

I remember teaching tuition on the first question

1) (B) ==> Higher frequency of hitting card so higher pitch
or if we can increase number of teeths also can

Cannot be fewer teeth

2) So

| x |

Time to travel from x to left wall and back = 0.6 s
Time to travel from x to right wall and back = 0.8 s => By this time, the echo from the left wall has travelled 0.2s towards the right wall.

Echo from left wall will take further 0.6s to reach x, and echo from right wall will also take 0.6s to travel from x to left wall and back

so answer is 0.8s + 0.6s = 1.4s

(B)

davidche SG

All i can do now . is concur

bonkysleuth SG

I remember teaching tuition on the first question

1) (B) ==> Higher frequency of hitting card so higher pitch
or if we can increase number of teeths also can

Cannot be fewer teeth

2) So

| x |

Time to travel from x to left wall and back = 0.6 s
Time to travel from x to right wall and back = 0.8 s => By this time, the echo from the left wall has travelled 0.2s towards the right wall.

Echo from left wall will take further 0.6s to reach x, and echo from right wall will also take 0.6s to travel from x to left wall and back

so answer is 0.8s + 0.6s = 1.4s

(B)

eh, i don't kind of understnad the explanation for the question on the walls. when they want to find the time taken to hear 2nd echo, is it the echo that is heard first after the first echo or something else?

skythewood SG

| x |

A B C

sound from B to A bounce back to B ===> 0.6s

sound from B to C bounce back to B ===> 0.8s.

with this pic in mind, solve the question. trace the movement of the sound with your finger.

davidche SG

I <--x--> (guy) <-------------y--------------> I

The second echo is when both echos combine.if the distance to teh left is x and to the right is y. And x is smalled than y...

The first echo was heard after the sound travelled 2 x. Second echo was heard after sound travel 2y. The first echo would continue right after the first echo was heard and bounce back to the person.(2x+2y) Second echo wld continue to the left and bounce back to the person (2y+2x) again.

time taken for sound to go thru 2x is 0.6 and 2y is 0.8. thus 2x+ 2y = 1.4

I love to explain by equations

skythewood SG

I remember teaching tuition on the first question

1) (B) ==> Higher frequency of hitting card so higher pitch
or if we can increase number of teeths also can

Cannot be fewer teeth

2) So

| x |

Time to travel from x to left wall and back = 0.6 s
Time to travel from x to right wall and back = 0.8 s => By this time, the echo from the left wall has travelled 0.2s towards the right wall.

Echo from left wall will take further 0.6s to reach x, and echo from right wall will also take 0.6s to travel from x to left wall and back

so answer is 0.8s + 0.6s = 1.4s

(B)

you mean 0.8 s.

edit: oh, i miss the extra arrow explaining the 0.2 s. my bad.

lifeisgood9989 SG

wats a cog wheel? :D

Craappiboy SG

turn at high speed cannot turn higher arh

and cog wheel is wheel of gear lo

NuckingFuts SG

yupz, its B,B.

faster speed=higher f=higher pitch

2nd one is just .6+.8 so easy -_-

actually i think thinner card also can xD