(A) The relationship between wave speed $(v)$,frequency $(n)$,and wavelength $(\lambda)$ is given by the formula: $v = n \lambda$. Rearranging for wav

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(A) The relationship between wave speed $(v)$,frequency $(n)$,and wavelength $(\lambda)$ is given by the formula: $v = n \lambda$. Rearranging for wavelength,we get: $\lambda = \frac{v}{n}$. Given: Frequency $(n)$ = $4.2 \, MHz = 4.2 \times 10^6 \, Hz$. Speed of sound $(v)$ = $1.7 \, km/s = 1.7 \times 10^3 \, m/s$. Substituting the values: $\lambda = \frac{1.7 \times 10^3}{4.2 \times 10^6} = \frac{1.7}{4.2} \times 10^{-3} \approx 0.4047 \times 10^{-3} \, m = 4.047 \times 10^{-4} \, m$. Rounding to the nearest value,we get $\lambda \approx 4 \times 10^{-4} \, m$.

Class 11 Physics Waves and Sound Sound Waves (Longitudinal wave) and it’s Characteristics (Speed etc.)

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$A$ hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is $4.2 \, MHz$. The speed of sound in a tissue is $1.7 \, km/s$. The wavelength of sound in the tissue is close to:

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A
$4 \times 10^{-4} \, m$
B
$8 \times 10^{-3} \, m$
C
$4 \times 10^{-3} \, m$
D
$8 \times 10^{-4} \, m$

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Sound Waves (Longitudinal wave) and it’s Characteristics (Speed etc.)

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$A$ hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is $4.2 \, MHz$. The speed of sound in a tissue is $1.7 \, km/s$. The wavelength of sound in the tissue is close to:

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$A$ man standing on a cliff claps his hands and hears the echo after $1 \, s$. If the sound is reflected from another mountain and the velocity of sound in air is $340 \, m/s$,then the distance between the man and the reflection point is ..... $m$.

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Explain the propagation of sound waves in:
$(i)$ air
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