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Brrunno [24]
3 years ago
7

A simple pendulum consists of a brass sphere of mass m = 1 kg suspended on a string of length L ≈ 1 meter. The pendulum oscillat

es with amplitude A = 1 cm and period T = 2 seconds. Which of the following alterations will change the period of the pendulum’s oscillations to T ′ = 1 second? 1. Using a lighter brass sphere of mass m ′ ≈ 0.25 kg. 2. Using a shorter string of length L ′ ≈ 0.25 meters. 3. Using a longer string of length L ′ ≈ 2 meters. 4. Using a longer string of length L ′ ≈ 4 meters. 5. Using a shorter string of length L ′ ≈ 0.5 meters. 6. Increasing the oscillation amplitude to A ′ = 4 cm. 7. Using a heavier brass sphere of mass m ′ ≈ 4 kg. 8. Reducing the oscillation amplitude to A ′ = 0.25 cm.
Physics
1 answer:
hram777 [196]3 years ago
6 0

Answer:

2. Using a shorter string of length L ′ ≈ 0.25 meters

5. Using a shorter string of length L ′ ≈ 0.5 meters

Explanation:

The period of a pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

We see from the formula that the period of the pendulum depends only on its length, not on its mass or its amplitude of ocillation. Therefore, the only alterations that can change the period of the pendulum are the ones where its length is changed.

Moreover, we notice that the period is proportional to the square of the length: this means that in order to decrease the period of the pendulum (the problem asks us which alterations will reduce the period of the pendulum from 2 s to 1 s), the length of the pendulum should also be reduced.

Therefore, the only alterations that will reduce the period of the pendulum are:

2. Using a shorter string of length L ′ ≈ 0.25 meters

5. Using a shorter string of length L ′ ≈ 0.5 meters

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V_0=\frac{4}{3} \pi \, R^3\\V_0=\frac{4}{3} \pi \, (0.12\,cm)^3\\V_0=0.007238\,cm^3

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\Delta V = V_0\,\, \alpha_V\,\,\Delta C\\\Delta V = 0.007238229\, cm^3\,(0.000182)\,(36)\\\Delta V=0.0000474248\, cm^3

Now, we can use this difference in volume, to estimate the height of the cylinder of mercury with diameter 0.0045 cm (radius r= 0.00225 cm):

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4 years ago
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Explanation:

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s=ut+\dfrac{1}{2}at^2

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