Mr. Doppler in a circle

Classical Mechanics Level 5

A tuning fork is moving in clockwise direction on a circle centered at origin of radius 2 m \sqrt { 2 }m with speed 125 m / s 125m/s . Find the coordinates ( a , b ) (a,b) of the tuning fork when the stationary observer at ( 1 , 3 ) (1,3) hears maximum frequency.

Select a 6 + b 6 {a}^{6}+{b}^{6} .

D e t a i l s Details

Velocity of sound = v s = 1000 π { v }_{ s }=\frac { 1000 }{ \pi }

This problem is originally part of set Mechanics problems by Abhishek Sharma .
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8 4 5 3 7 2 6 1

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Abhishek Sharma by Abhishek Sharma , 22, India

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2 solutions

Ayush Verma
May 8, 2015

Follow these steps,

1. F i n d t h e p o s i t i o n o f f o r k w h e n p a t h o f s o u n d w a v e i s t a n g e n t i a l t o c i r c l e a n d t o w a r d s t h e o b s e r v e r ( n o t a w a y f r o m o b s e r v e r ) i t w i l l b e ( − 1 , 1 ) b u t t h i s i s n o t ( a , b ) a s ( a , b ) i s p o s i t i o n w h e n o b s e r v e r h e a r s t h e s o u n d . ( a n s w e r V w i l l n o t b e ( − 1 ) 6 + ( 1 ) 6 = 2 ) 2. F i n d t i m e t a k e n b y w a v e t o r e a c h t o o b s e r v e r b y d i v i d i n g d i s t a n c e b / w ( − 1 , 1 ) & ( 1 , 3 ) b y v e l o c i t y o f s o u n d . t = ( 3 − 1 ) 2 + ( 1 + 1 ) 2 1000 / π = π 2 500 s e c 3. F i n d a n g l e c o v e r e d b y f o r k d u r i n g t i m e t , θ = ω t = v r t = 125 2 × π 2 500 = π 4 L e t ( a , b ) = ( 2 cos ⁡ α , 2 sin ⁡ α ) α = 3 π 4 − π 4 = π 2 ( a , b ) = ( 2 cos ⁡ α , 2 sin ⁡ α ) = ( 0 , 2 ) ⇒ a 6 + b 6 = 0 + 8 = 8 1.Find\quad the\quad position\quad offork\quad when\quad path\quad ofsound\quad wave\quad is\quad \\ \\ tangential\quad to\quad circle\quad and\quad towards\quad the\quad observer\\ \\ (not\quad away\quad from\quad observer)\quad it\quad will\quad be\quad (-1,1)\quad but\quad this\quad is\quad \\ \\ not\quad (a,b)\quad as\quad (a,b)\quad is\quad position\quad when\quad observer\quad hears\quad \\ \\ the\quad sound.\quad (answerVwill\quad not\quad be\quad (-1)^{ 6 }+(1)^{ 6 }=2\quad )\\ \\ 2.Find\quad time\quad taken\quad by\quad wave\quad to\quad reach\quad to\quad observer\quad by\quad \\ \\ dividing\quad distance\quad b/w\quad (-1,1)\& (1,3)\quad by\quad velocity\quad of\quad sound.\\ \\ t=\cfrac { \sqrt { { \left( 3-1 \right) }^{ 2 }+{ \left( 1+1 \right) }^{ 2 } } }{ 1000/\pi } =\cfrac { \pi \sqrt { 2 } }{ 500 } sec\\ \\ 3.Find\quad angle\quad covered\quad by\quad fork\quad during\quad time\quad t,\\ \\ \theta =\omega t=\cfrac { v }{ r } t=\cfrac { 125 }{ \sqrt { 2 } } \times \cfrac { \pi \sqrt { 2 } }{ 500 } =\cfrac { \pi }{ 4 } \\ \\ Let\quad \left( a,b \right) =\left( \sqrt { 2 } \cos { \alpha } ,\sqrt { 2 } \sin { \alpha } \right) \\ \\ \alpha =\cfrac { 3\pi }{ 4 } -\cfrac { \pi }{ 4 } =\cfrac { \pi }{ 2 } \\ \\ \left( a,b \right) =\left( \sqrt { 2 } \cos { \alpha } ,\sqrt { 2 } \sin { \alpha } \right) =\left( 0,\sqrt { 2 } \right) \\ \\ \Rightarrow { a }^{ 6 }+{ b }^{ 6 }=0+8=8\\

5 Helpful 0 Interesting 0 Brilliant 0 Confused

I misread the question, and thought (-1,1 ) was the answer :(.

A Former Brilliant Member - 5 years, 7 months ago

My calculations !!!! argh ! :(

Keshav Tiwari - 6 years, 1 month ago

Nicely done.

Abhishek Sharma - 6 years, 1 month ago

I over-thought this problem, and so I spent way too long on it, only to get the wrong answer. The original level is too high, and so it's misleading :/ Otherwise, I would have solved this almost instantly without having to do complicated calculations.

Jake Lai - 6 years, 1 month ago

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Many people selected 2 as the answer which led to increase in the rating of this problem. I agree that it is not a Level 5 problem.

Abhishek Sharma - 6 years ago

Why doesnt my ans came exactly 8?

I did like this.

For the position of tangent , the point of contact is given by ( a , b ) (a,b) .

So We know, equation of tangent = y = m x + a 1 + m 2 y= mx+a\sqrt{1+m^2}

⟹ 3 = m + 2 + 2 m 2 \implies 3=m+\sqrt{2+2m^2}

⟹ 9 + m 2 − 6 m = 2 + 2 m 2 \implies 9+m^2-6m=2+2m^2 \

⟹ m 2 + 6 m − 7 = 0 \implies m^2+6m-7=0

⟹ ( m − 1 ) ( m + 7 ) = 0 \implies (m-1)(m+7)=0

By drawing the figure we will know that we need an equation with Negetive slope but positive intercepts,

⟹ m = − 7 ⟹ y = − 7 x + 2 1 + 49 \implies m=-7 \implies y=-7x+2\sqrt{1+49}

⟹ y = − 7 x + 10 \implies y=-7x+10

⟹ y 2 = ( 10 − 7 x ) 2 \implies y^2=(10-7x)^2

⟹ x 2 + ( 10 − 7 x ) 2 = 2 \implies x^2+(10-7x)^2=2

⟹ 25 x 2 − 70 x + 49 = 0 \implies 25x^2-70x+49=0

⟹ x = 7 5 , y = 1 5 \implies x=\dfrac{7}{5} , y=\dfrac{1}{5}

⟹ x 6 + y 6 = 7 6 + 1 5 6 \implies x^6+y^6=\dfrac{7^6+1}{5^6}

≈ 8 \approx \boxed{8}

Why do i need to approximate in this case?

Md Zuhair - 3 years, 4 months ago

The clue is: the path of the sound wave to the observer must be tangent to the circular path to attain maximum frequency.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Brilliant question

VANSH SARDANA - 1 year, 5 months ago

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