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3 years ago

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A 16 kg mass suspended from a light spring is replaced by a 4 kg mass. What factor changes the frequency of the oscillation? (a)

1/4
(b) 2.0
(c) 4.0
(d) 1/2
(e) does not change

1 answer:

AnnZ [28]3 years ago

5 0

Answer:

Frequency change by a factor of 2.

(b) is correct option.

Explanation:

Given that,

Mass = 16 kg

Replaced mass = 4 kg

We need to calculate the frequency

Using formula of frequency

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Put the value into the formula

$f_{1}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{16}}$

$f_{1}=\dfrac{1}{2\pi}\dfrac{\sqrt{k}}{4}}$ ....(I)

$f_{2}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{4}}$

$f_{2}=\dfrac{1}{2\pi}\dfrac{\sqrt{k}}{2}}$ ...(II)

$f_{2}=2f_{1}$

Hence, Frequency change by a factor of 2.

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From equation (3) , we get

$\lambda=\frac{hc}{E} \\\\\lambda=\frac{ 4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm$

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