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Korvikt [17]
3 years ago
11

An undamped 1.48 kg horizontal spring oscillator has a spring constant of 35.4 N/m. While oscillating, it is found to have a spe

ed of 3.90 m/s as it passes through its equilibrium position. What is its amplitude A of oscillation
Physics
1 answer:
givi [52]3 years ago
5 0

Answer: 0.798 m

Explanation:

Given

Mass of the spring oscillator, m = 1.48 kg

Force constant of the spring, k = 35.4 N/m

Speed of oscillation, v = 3.9 m/s

Kinetic Energy = 1/2 mv²

Kinetic Energy = 1/2 * 1.48 * 3.9²

KE = 0.5 * 22.5108

KE = 11.26 J

Using the law of conservation of Energy. The Potential Energy of the system is equal to Kinetic Energy of the system

KE = PE

PE = 1/2kx²

11.26 = 1/2 * 35.4 * x²

11.26 = 17.7x²

x² = 11.26 / 17.7

x² = 0.6362

x = √0.6362

x = 0.798 m

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A certain thin lens is made of glass with refraction index ????lens=1.500. In air, where the index of refraction is 1.000, the l
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Using formula of focal length for glass air system

\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})

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\dfrac{1}{f'}=(n_{g}-n_{ethyl})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})

Divided equation (II) by (I)

\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}

Where, n_{g} = refractive index of glass

n_{a} = refractive index of air

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Put the value into the formula

\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}

\dfrac{f'}{11.5}=\dfrac{25}{7}

f'=\dfrac{25}{7}\times11.5

f'=41.07\ cm

Hence, The focal length of the lens in ethyl alcohol is 41.07 cm.

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