During an earthquake, where is the greatest motion felt on the surface?
1 answer:
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Answer:
c)by a factor of four
Explanation:
The total energy of a simple harmonic oscillator is given by
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
So, the total energy increases by a factor 4.
Answer:
Explanation:
Let the extension in the spring be x .
restoring force = weight of block
kx = mg
x =
= 23.84 cm
b )
When the elevator is going upwards
Restoring force = mg + ma
k x₁ = 10.9 ( 9.8 + 1.89 )
x₁ = 28.44 cm
( y coordinate will be - ( 28.44 - 23.84 ) = - 4.6 cm )
c ) When the cable snaps , both elevator and block undergo free fall . In this case apparent g = 0
Since the spring is stretched by 28.44 cm , a restoring force continues to act on the block which is equal to
.2844 x 448
= 127.41 N
So a net acceleration a will act on the block
a = 127.41 / 10.9
= 11.68 m / s²
The block will undergo SHM with amplitude equal to 28.44 cm .