Question

Two tuning Forks having frequencies 320 Hz
and 340 Hz ore Sounded together to produce
sound waves the velocity of sound in air
is 326.4 m/s Find the difference in
wavelength of these waves
​

Answer 1

Given:

1st Tuning Fork

Frequency (f1) = 320Hz

Velocity of sound in air (v) = 326.4m/s

2nd Tuning Fork

Frequency (f2) = 340Hz

Velocity of sound in air (v) = 326.4m/s

To find the difference in wavelength of the waves produced by them:

Solution:

Wavelength (λ1) of the 1st tuning fork is,

λ1 = v/f1

=326.4/320

λ1 = 0.98m

Wavelength (λ2) of the 2nd tuning fork is,

λ2 = v/f2

= 326.4/340

λ2=0.96m

Difference of these wavelengths = λ1-λ2 =0.98-0.96 =0.02m

So, the difference in wavelength of these waves is 0.02m.

Answer 2

Answer:

Concept:

The length of a complete wave cycle is measured in wavelength. A wave's velocity is the distance traveled by a point on the wave. In general, the relationship between Velocity and Wavelength is proportional for any wave.

Explanation:

wavelength and Velocity

The velocity of a wave is given by the product of its wavelength and frequency. It is given mathematically by the wave velocity formula, which is written as

$V = f \times \lambda$

Where,

V is the velocity of the wave measured in meters per second.

f  is the frequency of the wave measured in hertz.

$\lambda$  is the wavelength of the wave measured in millimeters.

To determine the wavelength difference between the waves produced by them:

Solution:

The first tuning fork's wavelength is,

$\lambda1$ = v/f1

    =326.4/320

$\lambda1$ = 0.98m

The second tuning fork's wavelength  is,

$\lambda2$ = v/f2

    = 326.4/340

$\lambda2$  =0.96m

The difference between these wavelengths =0.98 – 0.96 = 0.02m

So the wavelength difference between these waves is 0.02m.

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