Question:
A source of sound producing wavelength of $50 \text{ cm}$ is moving away from stationary observer with $\frac{1}{5}\text{th}$ speed of sound. The wavelength of the sound heard by the observer is
A source of sound producing wavelength of $50 \text{ cm}$ is moving away from stationary observer with $\frac{1}{5}\text{th}$ speed of sound. The wavelength of the sound heard by the observer is
Show Hint
If a source moves away, the wavelength always increases (redshift-like behavior in sound). If it moves towards, the wavelength decreases. This helps you immediately eliminate option (C).
Updated On: Jul 1, 2026
- $70 \text{ cm}$
- $55 \text{ cm}$
- $40 \text{ cm}$
- $60 \text{ cm}$
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
This problem is an application of the Doppler Effect for sound, specifically regarding the change in apparent wavelength when a source moves away from an observer.
Step 1: Identify the Wavelength Formula
When a source moves away from a stationary observer, the apparent wavelength $\lambda'$ is given by:
$$\lambda' = \lambda \left( \frac{v + v_s}{v} \right)$$
Where:
• $\lambda = 50 \text{ cm}$ (original wavelength)
• $v$ = speed of sound
• $v_s = \frac{1}{5}v$ (speed of source)
Step 2: Substitute and Solve $$\lambda' = 50 \left( \frac{v + \frac{1}{5}v}{v} \right)$$ $$\lambda' = 50 \left( \frac{\frac{6}{5}v}{v} \right)$$ $$\lambda' = 50 \times \frac{6}{5}$$ $$\lambda' = 10 \times 6 = 60 \text{ cm}$$ The wavelength heard by the stationary observer is $60$ cm.
• $\lambda = 50 \text{ cm}$ (original wavelength)
• $v$ = speed of sound
• $v_s = \frac{1}{5}v$ (speed of source)
Step 2: Substitute and Solve $$\lambda' = 50 \left( \frac{v + \frac{1}{5}v}{v} \right)$$ $$\lambda' = 50 \left( \frac{\frac{6}{5}v}{v} \right)$$ $$\lambda' = 50 \times \frac{6}{5}$$ $$\lambda' = 10 \times 6 = 60 \text{ cm}$$ The wavelength heard by the stationary observer is $60$ cm.
Was this answer helpful?
0
0
Top AP ECET EIE Physics Questions
- If the unit of mass is $1\text{ kg}$, the unit of length is $1\text{ m}$ and the unit of time is $1\text{ minute}$, the unit of pressure in $\text{Nm}^{-2}$ is
- $MLT^{-1}$ is the dimensional formula for
- If $|\vec{A} \times \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}$ then the value of $|\vec{A} + \vec{B}|$ is
- Of the vectors given below, the parallel vectors are $\vec{A} = 6\hat{i} + 8\hat{j}$, $\vec{B} = 210\hat{i} + 280\hat{k}$, $\vec{C} = 5.1\hat{i} + 6.8\hat{j}$, $\vec{D} = 3.6\hat{i} + 8\hat{j} + 48\hat{k}$
- The position $x$ of a particle with respect to time '$t$' along x- axis is given by $x = 9t^2 - t^3$ where $x$ is in metres and $t$ in seconds. The position of this particle when it achieves maximum speed along the x direction is
View More Questions
Top AP ECET EIE Traveling Sound Waves Questions
Top AP ECET EIE Questions
- Order of the matrix $\begin{bmatrix} 1 & 6 \\ 2 & 0 \\ 7 & -1 \end{bmatrix}$ is
- If two rows (or columns) of a determinant of order 3 are identical then the value of determinant is
- Co-factor of -4 in $\begin{vmatrix} 1 & 2 & 3 \\ -4 & 3 & 6 \\ 2 & -7 & 9 \end{vmatrix}$ is
- The Matrix $\begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$ is
- If $A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$ then $(A^{-1}) =$
View More Questions