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Ulleksa [173]
3 years ago
13

How should the mass of a harmonic oscillator be changed to double the frequency? Can the frequency be tripled by a suitable adju

stment of the mass?
Physics
1 answer:
Mars2501 [29]3 years ago
5 0

Answer:

a. m' = \frac{m}{4}

b. m' = \frac{m}{9}

Explanation:

The frequency of a harmonic oscillator is given by the following formula:

\omega = \sqrt{\frac{k}{m}}   ----------------- equation (1)

a.

In order to double the frequency of this oscillator:

ω' = 2ω

m' = ?

Therefore,

\omega ' = 2\omega = \sqrt{\frac{k}{m'}}

using equation (1):

2 \sqrt{\frac{k}{m}} = \sqrt{\frac{k}{m'}}\\\\ \frac{4}{m} = \frac{1}{m'}  

m' = \frac{m}{4}

a.

In order to triple the frequency of this oscillator:

ω' = 3ω

m' = ?

Therefore,

\omega ' = 3\omega = \sqrt{\frac{k}{m'}}

using equation (1):

3\sqrt{\frac{k}{m}} = \sqrt{\frac{k}{m'}}\\\\ \frac{9}{m} = \frac{1}{m'}  

m' = \frac{m}{9}

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On a hypothetical scale X The ice point is 40° and steam point is 120°.
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Answer:

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3 years ago
A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c
den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

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T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

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L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

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