The two speakers of a boom box are 37.8 cm apart. A single oscillator makes the

Question

The two speakers of a boom box are 37.8 cm apart. A single oscillator makes the speakers vibrate in phase at a frequency of 1.82 kHz. At what angles, measured from the perpendicular bisector of the line joining the speakers, would a distant observer hear maximum sound intensity? (Take the speed of sound as 340 m/s. Consider nonnegative angles only. Enter your answers from smallest to largest. Enter NONE in any unused answer blanks.)

There are 4 answers

At what angles would such an observer hear minimum sound intensity? (Take the speed of sound as 340 m/s. Consider nonnegative angles only. Enter your answers from smallest to largest. Enter NONE in any unused answer blanks.)

There are 3 answers

Explanation / Answer

Given that

The two speakers of a boom box are (e) =37.8 cm apart =0.378m

A single oscillator makes the speakers vibrate in phase at a frequency of (f) =1.82 kHz =1820Hz

The speed of sound is given by (v) =340m/s

The condition for maximum intensity is given by

esintheta =mlamda

theta =sin-1(mlamda/e)

We also know that v=f*lamda then lamda=v/f =340m/s/1820Hz =0.1868m

When m=0

Then the angle theta =sin-1(m*0.1868m/0.378m)=0

When m=1

Then the angle theta =sin-1(1*0.1868m/0.378m)=29.617degrees

Similalry the condition for the minimum intensity is given by

esintheta =(m+(1/2))lamda

Then theta =sin-1((m+(1/2))lamda/e)

When m =0

theta =sin-1((m+(1/2))lamda/e)=(0.1868m/2*0.378m) =14.305degrees

When m=1

then theta =sin-1((m+(1/2))lamda/e)=(3*0.1868m/2*0.378m) =47.839degrees

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.