The ___of a wave is the number of wavelengths that pass a fixed point in a second.

Physics

1 answer:

creativ13 [48]3 years ago

7 0

the frequency of a wave is the number of wavelengths that pass a fixed point in a second.

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A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What wil

mash [69]

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

$F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}$

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

$\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\$

when the tension is decreased to 600 N, that is T₂ = 600 N

$T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm$

Therefore, the wavelength will be 33.9 cm

5 0

3 years ago

What is a concave lens?

ozzi

A concave lens is a lens that has at least one of its surfaces or both surfaces curved inwards. Due to this reason, this lens diverges the light that falls on it and hence is also called a diverging lens. The concave lens is thinner in the middle compared to its edges. These are used in flashlights, binoculars, telescopes, etc.
Please see attached image for reference.