Two loudspeakers L1 and L2 driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is 330 ms–1 then the frequency at which the first maximum is observed is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

165 Hz

330 Hz

496 Hz

660 Hz

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Path difference between the wave reaching at D Δx=L2P−L1P=402+92−40=41−40=1mFor maximum Δx=(2n)λ2For first maximum n=1⇒1=2(1)λ2⇒λ=1m⇒n=vλ=330Hz

Watch 3-min video & get full concept clarity