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kaheart [24]
3 years ago
9

Imagine a simple pendulum swinging in an elevator. If the cable holding the elevator up was to snap, allowing the elevator to go

into free fall, what would happen to the frequency of oscillation of the pendulum? Justify your answer.
Physics
1 answer:
Lady_Fox [76]3 years ago
5 0

One of the components that affect the period is gravity (the other is length). This gravity is basically the value of the effective acceleration that acts on the body due to gravity. When the elevator is over free fall, the effective gravity becomes zero. Mathematically this can be visualized as,

T = 2\pi \sqrt{\frac{l}{\dot{g_{eff}}}}

Since this value is zero, the period would tend to be infinite,

T \rightarrow \infty

Therefore the frequency that is inversely proportional to the period would be defined as

f = \frac{1}{T}

f = \frac{1}{\infty}

f \approx 0

In this way there is no frequency on the body which will not generate any oscillation on the body

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A spring with a mass on the end of it hangs in equilibrium a distance of 0.4200 m above the floor. The mass is pulled down a dis
den301095 [7]

Answer:

0.48 m

Explanation:

I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others

If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.

0.42 m + 0.06 m = 0.48 m

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3 0
3 years ago
A 20-cm-long, 190 g rod is pivoted at one end. A 19 g ball of clay is stuck on the other end.
MaRussiya [10]

Answer:

0.76 s

Explanation:

We are given that

Length of rod,L=20 cm=\frac{20}{100}=0.20m

1 m=100 cm

Mass of rod,M=190 g=\frac{190}{1000}=0.19kg

Mass of ball,m=19 g=\frac{19}{1000}=0.019 kg

Using 1 kg=1000g

We have to find the period if the rod and clay swing as a  pendulum.

Moment of inertia of rod-clay=Moment of inertia of rod+moment of inertia of clay

I_{rod-clay}=I_{rod}+I_{clay}

I_{rod-clay}=\frac{1}{3}ML^2+mL^2

Substitute the values then we get

I_[rod-clay}=\frac{1}{3}(0.19)(0.20)^2+(0.019)(0.20)^2

I_{rod-clay}=3.29\times 10^{-3} kgm^2

Now, the center of mass of the combination of the rod and clay is given by

y=\frac{Md_1+md_2}{M+m}

Substitute d_1=\frac{L}{2}=Distance between pivot and the center of the rod

d_2=L=The distance  between rod and clay

Using the formula

y=\frac{0.19\times \frac{0.20}{2}+0.019\times 0.20}{0.19+0.019}

y=0.1091 m

Time period of the oscillation of the system of the rod and the clay is given by

T=2\pi\sqrt{\frac{I_{rod-clay}}{(M+m)yg}}

g=9.8m/s^2

Using the formula

Time period=2\times 3.14\sqrt{\frac{3.29\times 10^{-3}}{(0.19+0.019)\times 9.8\times 0.1091}}

Time-period=0.76 s

Hence, the period =0.76 s

8 0
3 years ago
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