A system of four particles moves along a dimension. The center of mass is at rest, and the particles do not interact with any ob
jects outside of the system. Find the velocity of v4 at t=2.83 seconds given the details for the motion of particles 1,2,3
1 answer:
Answer:
v = - 14.08 m / s
Explanation:
The definition of center of mass is
= 1 /M ∑sun
where M is the total mass of the system and and
are the position and mass of each component.
The velocity of the center of mass can be found by deriving this expression with respect to time
= 1 / M ∑ m_{i}
vi
let's find the total mass
M = m₁ + m₂ + m₃ + m₄
M = 1.45 + 2.81 +3.89 + 5.03
m = 13.18 kg
let us substitute in the velocity of the center of mass = 0
0 = 13.18 (m₁ v₁ + m₂ v₂ + m₃v₃ + m₄v₄)
v₄ = - (m₁ v₁ + m₂ v₂ + m₃v₃) / m₄
let's substitute the given values
v₄ = -[1.45 (6.09 +0.299 t) +2.81 (7.83 + 0.357t) +3.89 (8.09 + 0.405 t)] / 5.03
They ask us for the calculations for a time t = 2.83 s
v₄ = - [8.8305 + 1.227 + 22.00 + 2.839 + 31.47 +4.4585] / 5.03
v = - 14.08 m / s
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Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,
Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀ =A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values
b)
, where
is time period of damped SHM
⇒
let be number of oscillations made
then,
⇒
1) The length of the lead is 1700 m
2) The frequency of the wave is 400 Hz
Explanation:
1)
The wave in this problem is a sound wave, which is produced by the vibrations of the particles of air. It is a type of longitudinal wave (which means that the vibrations occur back-and-forth along the direction of propagation of the wave).
The speed of a sound wave in air is approximately
v = 340 m/s
In this problem, the sound wave is heard after a time of
t = 5 s
after the shot. Therefore, the distance travelled by the wave in this time (and therefore, the length of the lead) is
2)
The frequency of a wave is the number of complete cycles made by the wave in one second.
For the wave in this problem, we are told that the wave has 2000 whole wavelengths in 5 seconds: this means that the wave completes 2000 cycles in 5 seconds. Therefore, we can find the frequency by setting up the following proportion:
where x is the number of cycles made in 1 second. Solving for x,
Therefore, the frequency of the wave is
Learn more about waves and frequency:
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