Answer:
The frequency would double.
Explanation:
Given:
Speed of wave (v) = constant.
Frequency of wave initially (f₁) = 2 Hz
Initial wavelength of the wave (λ₁) = 1 m
Final wavelength of the wave (λ₂) = 0.5 m
Final frequency of the wave (f₂) = ?
We know that the product of wavelength and frequency of the wave is equal to the speed of the wave.
Therefore, framing in equation form, we have:
Wavelength × Frequency = Speed
[tex]\lambda\times f=v[/tex]
It is given that speed of the wave remains the same. So, the product must always be a constant.
Therefore,
[tex]\lambda\times f=constant\ or\ \\\lambda_1\times f_1=\lambda_2\times f_2[/tex]
Now, plug in the given values and solve for 'f₂'. This gives,
[tex]1\times 2=0.5\times f_2\\\\f_2=\frac{2}{0.5}=4\ Hz[/tex]
Therefore, the final frequency is 4 Hz which is double of the initial frequency.
f₂ = 2f₁ = 2 × 2 = 4 Hz
So, the second option is correct.
