A stone is rotated steadily in a horizantal circle with a period T by a string of length L. If the tension in the stringis constant and L increases by 1% find the percentage change in T.
centripetal acceleration = a = v^2/L
Period = T = 2 x pi x L / v
m = mass of stone
tension = f = m x a
L1 = L x 1.01
if f and m are constant, then a is constant. Then
v1 = sqrt(a x L1)
T1 = 2 x pi x L1/v1
percentage change = 100x(T1-T)/T
To find the percentage change in the period T when the length L increases by 1%, we need to use the concept of simple harmonic motion.
In simple harmonic motion, the period T (time taken for one complete revolution) of an object moving in a circular path is given by:
T = 2π√(L/g),
where L is the length of the string and g is the acceleration due to gravity.
Now, let's calculate the change in T when L increases by 1%.
1. First, find the initial period T1 for the initial length L:
T1 = 2π√(L/g).
2. Next, find the final period T2 for the increased length L + 1%:
T2 = 2π√((L + 0.01L)/g)
= 2π√((1.01L)/g)
= 2π√(1.01L/g).
3. Calculate the percentage change in T by using the formula:
Percentage change = ((T2 - T1) / T1) * 100.
Substituting the values of T1 and T2 into the formula, we get:
Percentage change = (((2π√(1.01L/g)) - (2π√(L/g))) / (2π√(L/g))) * 100
= (((√(1.01L/g)) - √(L/g)) / √(L/g)) * 100.
Simplifying further:
Percentage change = ((√(1.01L/g) - √(L/g)) / √(L/g)) * 100.
Thus, the percentage change in T when L increases by 1% is given by ((√(1.01L/g) - √(L/g)) / √(L/g)) * 100.