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3 years ago

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If a vibrating string is made shorter (i.e., by holding your finger on it), what effect does this have on the frequency of vibra

tion and on the pitch? 1. both decrease. 2. no effect 3. both increase. 4. the frequency decreases; the pitch decreases. 5. the frequency increases; the pitch decreases.

1 answer:

Temka [501]3 years ago

4 0

The correct answer is "3. Both increase"

In fact, the frequency of a vibrating string is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where L is the length of hte string, T the tension and $\mu$ the linear mass density.
We can see that string is made shorter, L decreases, so the frequency f increases. And since the pitch has the same behaviour of the frequency, it increases as well.

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3 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

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As we know that KE and PE is same at a given position

so we will have as a function of position given as

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also the PE is given as function of position as

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now it is given that

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now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

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Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

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$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

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3 years ago

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Answer:

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3 years ago

Read 2 more answers

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Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity $u=0$

acceleration $a=12.8 m/s^2$

velocity acquired by sled in $t_1$ time

$v=0+at$

$v=12.8t_1$

distance traveled by sled in $t_1 s$

$v^2-u^2=2as$

$(12.8t_1)^2-0=2\times 12.8\times s_1$

$s_1=6.4\cdot t_1^2$

distance traveled in $t_2$ time with velocity $v=12.8t_1$

$s_2=v\times t_2$

$s_2=12.8\times t_1\times t_2$

$s_2=12.8\cdot t_1\cdot t_2$

$s_1+s_2=5.37\times 10^3$

$6.4t_1^2+12.8t_1t_2=5370$ ----1

$t_1+t_2=97.7 s$

$t_2=97.7-t_1$

substitute the value of $t_2$ in 1

we get

$6.4t_1^2-1250.56t_1+5370=0$

thus $t_1=\frac{1250.56-1194.33}{12.8}=4.39 s$

$t_1=4.39 s$

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