Two heavy bodies of mass 'm' and '3m' tied together with a
light string of mass density u are dropped from a helicopter
as shown. A constant air friction force 'F' acts on both of
them. If the length of the string is (, determine the time
taken by a pulse to go from top to bottom of the string.
m
3m.
Answers
Thus the formula for time is t = 2 π √ l / g
Explanation:
- Mass of body m1 = m
- Mass of second body m' = 3 m
- Force of friction = F
- To Find: Time to go from top to bottom
Solution:
As we know that for Simple Harmonic motion.
The formula of time period is given as.
t = 2 π √ l / g
Where t = time
π = 22/ 7
L = length of pendulum / string
g = gravitational acceleration
Thus the formula for time is t = 2 π √ l / g
The time taken by a pulse to go from top to bottom of the string is l × √((2μ)/F)
Explanation:
The force acting on mass 'm' is given as:
T + mg - F = ma → (Equation 1)
The force acting on mass '3m' is given as:
3mg - (T + F) = 3ma → (Equation 2)
On multiplying 3 in equation (1), we get,
3T + 3mg - 3F = 3ma
Now,
3mg - (T + F) = 3T + 3mg - 3F
3mg - T - F = 3T + 3mg - 3F
3T + T - 3F + F = 0
4T - 2F = 0
On dividing above equation with 2, we get,
2T - F = 0
2T = F
∴ T = F/2
The velocity of the oscillation is given as:
v = √(T/μ)
On substituting 'T', we get,
∴ v = √(F/2μ)
Now, the time is given as:
Time = (Distance)/(Velocity) ⇒ t = D/v
D = l (The distance traveled by pulse is length of the string)
t = l/√(F/2μ)
∴ t = l × √(2μ/F)