Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound waves in phase at 688 Hz. Take the speed of sound in air to be 344 m/s. What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers?

Question:

If you are 3.00 m from speaker A directly to your right and 3.50 m from speaker B directly to your left. What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers?

The shortest distance d you need to walk forward to be at a point where you cannot hear the speakers is 5.625 m

Explanation:

Shortest distance is given by

For constructive interference of wave, we have

= 2nλ/2

For destructive interference of wave, we have

= (2n+1)λ/2

That is the difference in the number of wavelength between constructive and destructive interference is λ/2

We note that our positioning is 3.0 m from the first speaker and 3.5 m from the second, Which is of the form 2n and 2n + 1

Therefore when we walk forward away from the two speakers, we cannot hear the speakers at

Δr = λ/2 = r - r

Which is the distance between the points where we have constructive and destructive interference or where there is destructive interference between the waves

Where:

r =  Distance from  speaker A and

r = Distance from  speaker B

λ/2 = (344 m/s /688 Hz)/2

= 1/4

For

λ/2 = r - r  we have

√(3.5² +d²) - √(3²+d²) = 1/4

∴ d = 5.625 m

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards