Answer:
The frequency would double.
Step-by-step 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
![\lambda* f=v](https://img.qammunity.org/2021/formulas/physics/middle-school/qkp6dhtjpb1mocgbwwtyt8jvud42us520z.png)
It is given that speed of the wave remains the same. So, the product must always be a constant.
Therefore,
![\lambda* f=constant\ or\ \\\lambda_1* f_1=\lambda_2* f_2](https://img.qammunity.org/2021/formulas/physics/middle-school/h1lbcr5u359pk2cn8aqikieon5157fmg8k.png)
Now, plug in the given values and solve for 'f₂'. This gives,
![1* 2=0.5* f_2\\\\f_2=(2)/(0.5)=4\ Hz](https://img.qammunity.org/2021/formulas/physics/middle-school/4vafh1jxvezkmn0sqs60j8d856cgc47lrv.png)
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.