A heavy ball of mass M is suspended from the ceiling of car by a light string of mass m (m << M). When the car is at rest, the speed of transverse waves in the string is 60 ms⁻¹. When the car has acceleration a, the wave-speed increases to 60.5 ms⁻¹. The value of a, in terms of gravitational acceleration g is closest to:
(A) g/30
(B) g/5
(C) g/10
(D) g/20
[JEE Main 2019]
Answers
Thus the value of a is a = g / 5
Option (B) is correct.
Explanation:
Speed of wave on string -
v = √T / μ
Where;
T = tension
μ = linear mass density
v = √T / m
60 = √T / m since m < < M
Resultant acceleration = √a^2 + g^2
60.5 = √M √a^2 + g^2/m
(60.5)^2 = √a^2 + g^2/m
(60.5 / 60)^2 = a^2 + g^2/m
a^2 = g^2 x (60.5)^2 / 60^4 - g^2
a = g √(60.5 / 60)^4 - 1
a = g / 5
Thus the value of a is a = g / 5
The value of a in terms of gravitational acceleartion g is closest to g/5
given
heavy ball of mass M is suspended by a light string of mass m
now the velocity v=√T/μ
suppose length of string =L
then m/L=μ
the mass of two forces acting on it one is tension T(upwards) and the other is Mg downwards
by balancing forces
T=Mg
v=√Mg/μ=60.............(1)
now when car accelarates the wave speed is 60.5ms⁻¹
The mass now gets a pseudo force of Ma in the direction opposite to the car is moving
if the ball moves by angle θ
then T= Mgcos θ +Masin θ(in y axis)
Mgsin θ=Ma cos θ
tan θ=a/g
sin θ=a/√a²+g²,cos θ=g/√a²+g²
T=Mg*g/√a²+g² +Ma *a/√a²+g²
=M/√a²+g²(g²+a²)
T=M√g²+a²
therefore 60.5=√M√g²+a²/μ......................(2)
comparing 1 &2
60.5/60=√M√g²+a²/Mg
(1+0.5/60)^4=g²+a²/g²
0.5<<60
a=g/√30
=g/5