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3 years ago

If the amplitude of a simple harmonic oscillator is doubled, by what factor does the total energy increase?

Physics

1 answer:

dybincka [34]3 years ago

6 0

Answer:

c)by a factor of four

Explanation:

The total energy of a simple harmonic oscillator is given by

$E=\frac{1}{2}kA^2$

where

k is the spring constant of the oscillator

A is the amplitude of the motion

In this problem, the amplitude of the oscillator is doubled, so

A' = 2A

Therefore, the new total energy is

$E'=\frac{1}{2}k(2A)^2=4(\frac{1}{2}kA^2)=4E$

So, the total energy increases by a factor 4.

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A space vehicle accelerates uniformly from 85 m/s at t = 0 to 164 m/s at t = 10.0 s .How far did it move between t = 2.0 s and t

creativ13 [48]

First, we have a change in the velocity from 85 to 164 m/s in 10 sec.

Then, we calculate the <u>acceleration </u>as:

$a=\frac{v_{f}-v_{i} }{t} =\frac{164-85}{10}=7.9 m/s^2$

Hence we need to calculate the velocity of the space vehicle at t = 2 sec using the first equation of motion:

$v_{f}=v_{i}+at=85+7.9*2=100.8m/s$

Then, using the second equation of motion to calculate the distance:

$d=v_{i} t+\frac{1}{2}at^2$

$d=100.8*2+\frac{1}{2}*7.9*(2)^2=217.4m$

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3 years ago

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When summer changes to fall, what seasonal changes do plants experience? (IGNORE HIGHLIGHTED ANSWER)

lorasvet [3.4K]

Answer:

Explanation:

There is a decrease in temperature and daylight and plants produce less food.

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2 years ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

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3 years ago

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Eduardwww [97]

Answer:

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