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3 years ago

a waves amplitude is 0.5 meters. If the amplitude is increased to 1 metro, how does its energy change

Physics

1 answer:

marysya [2.9K]3 years ago

5 0

Answer:

The energy becomes 4 times greater.

Explanation:

We know that the energy of a wave is proportional to the square of its amplitude

E ∝ Amplitude^2

Since the original amplitude = 0.5 m

and the new amplitude becomes = 1 m

We are doubling the amplitude. This means that the new energy will be affected by a factor of 4

E_new ∝ (2*Amplitude)^2 =

E_new ∝ 4*(Amplitude)^2

E_new = 4*E

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The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of

IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_{beat} = 0.99s$

Generally the frequency of the beat is

$f_{beat} = \frac{1}{t_{beat}}$

Substituting values

$f_{beat} = \frac{1}{0.99}$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_{beat}$ for $f_2$ having a higher tension

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

$\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

$\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_{beat}$

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

$\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

$\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}$

$T_2 = 0.99$ % lower than $T_1$

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