Question

A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.0931 m, its frequency is 2.01 Hz,and its wavelength is 1.95 m.

What is the shortest transverse distance between a maximum and a minimum of the wave = _____m
How much time Δ is required for 77.3 cycles of the wave to pass a stationary observer?

Answer 1

The shortest transverse distance between a maximum and a minimum of a sinusoidal transverse wave is half its wavelength, and for 77.3 cycles at 2.01 Hz, the time required for them to pass a stationary observer is approximately 38.5 seconds.

Step-by-step explanation:

The question is about the characteristics of a sinusoidal transverse wave and the calculations related to its properties.

Shortest Transverse Distance Between a Maximum and a Minimum

For a sinusoidal wave, the shortest transverse distance between a maximum (crest) and a minimum (trough) is half the wavelength. Given the wavelength (λ) is 1.95 meters, the distance is simply λ/2, which equates to 0.975 m.

Time Required for 77.3 Cycles

To find the time ( Δ ) for 77.3 cycles for a wave with frequency (f) of 2.01 Hz, we use the formula: Δ = number of cycles / frequency. This gives Δ = 77.3 / 2.01 Hz, which results in approximately 38.5 seconds.