Question

A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.08190.0819 m, its frequency is 2.292.29 Hz, and its wavelength is 1.871.87 m. (a) What is the shortest transverse distance between a maximum and a minimum of the wave

Answer 1

Answer:

The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

Explanation:

Given that,

Amplitude = 0.08190 m

Frequency = 2.29 Hz

Wavelength = 1.87 m

(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave

Using formula of distance

$d=2A$

Where, d = distance

A = amplitude

Put the value into the formula

$d=2\times0.08190$

$d=0.1638\ m$

Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

Answer 2

The shortest transverse distance between a maximum and a minimum of the wave is 0.164 m.

Vertical distance between the crest and trough of the wave

The shortest transverse distance between a maximum and a minimum of the wave is the same as the vertical distance between the crest and trough of the wave.

The shortest transverse distance between a maximum and a minimum of the wave is calculated as follows;

d = 2A

where;

• A is the amplitude of the wave

d = 2 x 0.0819

d = 0.164 m

Thus, the shortest transverse distance between a maximum and a minimum of the wave is 0.164 m.

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