Answer:
is the distance from the obstacle of reflection.
wavelength
![\lamb=0.5279\ m](https://img.qammunity.org/2020/formulas/physics/high-school/vfdnuaknk5rtz9zoryjz02p89prn2mw7qw.png)
Step-by-step explanation:
Given that:
- frequency of sound,
![f=610\ Hz](https://img.qammunity.org/2020/formulas/physics/high-school/ldcatj6nzreng2m5i9ooabdlq5wmgncsoh.png)
- time taken for the echo to be heard,
![t=3.9\ s](https://img.qammunity.org/2020/formulas/physics/high-school/z498pdqfmds150rwlf08s6pd8d87h49kan.png)
- speed of sound,
![v=322\ m.s^(-1)](https://img.qammunity.org/2020/formulas/physics/high-school/74852a1kyuox5coomxtmlf8cgjrhq0edpg.png)
We know,
![\rm distance = speed * time](https://img.qammunity.org/2020/formulas/physics/high-school/wy0h5i1arg819duicgp1a4jwuhfda7cipa.png)
During an echo the sound travels the same distance back and forth.
![2d=v.t](https://img.qammunity.org/2020/formulas/physics/high-school/3trf5fn1gk9x3tlfnh53vqabguij8brphp.png)
![2d=322* 3.9](https://img.qammunity.org/2020/formulas/physics/high-school/vmiqoe7rgcynbjatit11mzrrge3uu4u7oc.png)
is the distance from the obstacle of reflection.
Now the wavelength of sound waves:
![\lambda=(v)/(f)](https://img.qammunity.org/2020/formulas/physics/high-school/3mt29el0otp87hhrg01vubljucwyu8v3a1.png)
![\lambda=(322)/(610)](https://img.qammunity.org/2020/formulas/physics/high-school/6zx116brqlc8dpjr705k0qv6si1wt7h820.png)
![\lamb=0.5279\ m](https://img.qammunity.org/2020/formulas/physics/high-school/vfdnuaknk5rtz9zoryjz02p89prn2mw7qw.png)