Respuesta :
For a simple harmonic motion energy is given with:
[tex]E=\frac{1}{2}kA^2[/tex]
Where k is a constant that depends on the type of the wave you are looking at and A is amplitude.
Let's calculate the energy of the wave using two different amplitudes given in the problem:
[tex]E_1=\frac{1}{2}k(1)^2=\frac{1}{2}k\\E_2=\frac{1}{2}k(2)^2=\frac{4}{2}k=2k\\[/tex]
We can see that energy associated with the wave is 4 times smaller when we decrease its amplitude by half. So the answer should be C.
[tex]E=\frac{1}{2}kA^2[/tex]
Where k is a constant that depends on the type of the wave you are looking at and A is amplitude.
Let's calculate the energy of the wave using two different amplitudes given in the problem:
[tex]E_1=\frac{1}{2}k(1)^2=\frac{1}{2}k\\E_2=\frac{1}{2}k(2)^2=\frac{4}{2}k=2k\\[/tex]
We can see that energy associated with the wave is 4 times smaller when we decrease its amplitude by half. So the answer should be C.
Answer:
It's C
Explanation:
Since E ∝ amplitude2 then as the amplitude is halved the energy is quartered.
Plugging in to check
amplitude2=E
22=4
12=1