in fig. 15.12, two identical springs of spring constant 7580 n/m are attached to a block of mass 0.245 kg. what is the frequency
1 answer:
The frequency of oscillation on the frictionless floor is 39.6Hz
Frequency is defined as the number of oscillations in unit time.
The net force applied by the springs to bring the block back to equilibrium when it is out of balance is -2kx acting in that direction.
We know that,
a = dx²/dt²
From Newton second law of motion,
F = ma
mdx²/dt² = -2kx
On substituting, x=x' cos(ωt+ϕ) where x' is the position at mean position.
We get, w² = 2k/m
Hence, w =
w =
w = 248.75
Since there are 2π radians in a cycle, and frequency f measures cycles per second,
f = w/2π
On substituting the value of w,
f = 248.75/2π
f = 39.6Hz
Hence, the frequency of oscillation on the frictionless floor is 39.6Hz
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Answer:
<em>I must travel with a speed of 2.97 x 10^8 m/s</em>
Explanation:
Sine the spacecraft flies at the same speed in the to and fro distance of the journey, then the time taken will be 6 months plus 6 months
Time that elapses on the spacecraft = 1 year
On earth the people have advanced 120 yrs
According to relativity, the time contraction on the spacecraft is gotten from
=
where
is the time that elapses on the spacecraft = 120 years
= time here on Earth = 1 year
is the ratio v/c
where
v is the speed of the spacecraft = ?
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
120 = 1/
squaring both sides of the equation, we have
14400 = 1/
14400 - 14400 = 1
14400 - 1 = 14400
14399 = 14400
= 14399/14400 = 0.99
= 0.99
substitute β = v/c
v/c = 0.99
but c = 3 x 10^8 m/s
v = 0.99c = 0.99 x 3 x 10^8 = <em>2.97 x 10^8 m/s</em>