Help me plz on the one i just put i neeed it uhg plz anter
1 answer:
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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,
⇒
Answer:
The required angle is (90-25)° = 65°
Explanation:
The given motion is an example of projectile motion.
Let 'v' be the initial velocity and '∅' be the angle of projection.
Let 't' be the time taken for complete motion.
Let 'g' be the acceleration due to gravity
Taking components of velocity in horizontal(x) and vertical(y) direction.
= v cos(∅)
= v sin(∅)
We know that for a projectile motion,
t =
Since there is no force acting on the golf ball in horizonal direction.
Total distance(d) covered in horizontal direction is -
d = ×t = vcos(∅)×
=
.
If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -
α +β = 90° as-
d = =
=
=
.
∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.
∴ If α = 25° , then
β = 90-25 = 65°
∴ The required angle is 65°.