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3 years ago

12

When tuning a guitar, a musician plays two strings at once. These sound waves overlap and often appear to get louder and softer

if the strings are not in tune. This interaction of overlapping waves is called

1 answer:

Ymorist [56]3 years ago

3 0

It's called ' interference '.

If the strings are 'in tune' (same frequency) and in phase, then
they overlap to make a louder sound ... CONstructive interference.

If the strings are 'in tune' (same frequency) but out of phase, then
they overlap to make a softer sound ... DEstructive interference.

If the strings are 'out of tune' (different frequencies), then they overlap
to make a sound that's louder at some times and softer at other times.
The louder and softer pattern creates a new sound, called the 'beat'.
Its frequency is the difference in the frequencies of the two strings.

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3 years ago

A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The

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Answer:

1. $v=14.2259\ m.s^{-1}$

2. $F_T=25.8924\ N$

3. $\lambda=29.6373\ m$

Explanation:

Given:

mass of slinky, $m=0.87\ kg$

length of slinky, $L=6.8\ m$

amplitude of wave pulse, $A=0.23\ m$

time taken by the wave pulse to travel down the length, $t=0.478\ s$

frequency of wave pulse, $f=0.48\ Hz=0.48\ s^{-1}$

1.

$\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel$

$v=\frac{6.8}{0.478}$

$v=14.2259\ m.s^{-1}$

2.

<em>Now, we find the linear mass density of the slinky.</em>

$\mu=\frac{m}{L}$

$\mu=\frac{0.87}{6.8}\ kg.m^{-1}$

We have the relation involving the tension force as:

$v=\sqrt{\frac{F_T}{\mu} }$

$14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }$

$202.3774=F_T\times \frac{6.8}{0.87}$

$F_T=25.8924\ N$

3.

We have the relation for wavelength as:

$\lambda=\frac{v}{f}$

$\lambda=\frac{14.2259}{0.48}$

$\lambda=29.6373\ m$

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3 years ago