I also have this solution:
The natural length of the upper and lower demarcations are 2 and 3 respectively. Let tension in upper string be T(1) and T(2) for the lower.
Resolving vertically (up as positive) and at equilibrium -
Elastic modulus = 6; Extension in upper string henceforth = (4 + x) and in lower = (6 - x);
T(1) = Weight + T(2)
6(4+x)/2 = 0.5*9.8 + 6(6-x)/3
12 + 3x = 12 - 2x + 4.9
5x = 4.9
Therefore, midpoint of oscillation is when x = 0.98;
This value of 0.98m is also the amplitude of the oscillation.
Resolving vertically (up as +ve) at any time t :
T(1) - mg - T(2) = m * d^2x/dt^2
6(4+x)/2 - 6(6-x)/3 - 4.9 = 0.5 * d^2x/dt^2
12 + 3x - 12 +2x - 4.9 = 0.5 * d^2x/dt^2
5x - 4.9 = 0.5 * d^2x/dt^2
take factor of -5 from LHS
-5(0.98 - x) = 0.5 * d^2x/dt^
d^2x/dt^2 = -10(0.98 - x)
let y = 0.98 - x
d^2y/dt^2 = -10y
we know that n = √k ; d^2y/dt^2 = - ky
Therefore, n = √10
Period of oscillation = 2*pi/n = 1.986... s
Since the amplitude is 0.98, the lamp will fall through a distance (2 x 0.98) = 1.96.
The lamp is dropped from height 9m; it will thus oscillate between 9m and (9-1.96)m = 7.04m
I hope this helps somebody else in the future.
See, basically the thing is tension in classical mechanical systems are used for balancing applied forces in a network of applied forces. Having said that it is also important to keep in mind that tension is a bi-directional force which means whatever you have heard like ------>---<----- is correct but it depends upon the situation.
As I said, a tension or a tensional force is always away from an object or a point in the system. To understand this, let us analyse the given system that you have mentioned....
Consider point A,
tension T1 (say) is acting through the string towards point R.
Consider point R,
tension T1 (bi-direction in nature) is acting towards point A.
Tension T2 (say) is acting towards point S.
Notice that T1 and T2 are perpendicular to eachother.
Consider point S,
Tension T3 say is acting towards 2g wt.
Tension T2 is acting towards the point R.
Now we will frame equations as per requirements and we will try to eliminate the tension forces from the equations, this is our objective for solving any mechanical problem.
In this case,
Look at the diagram and try to understand the following:
$T_3 = 0.002 kg wt.$
Again at point S,
$T_2 sin (\alpha) = T_3$
At point R,
$T_2 sin (\alpha) = F sin (\theta) + T_1 cos (\alpha)$
$T_1 sin (\alpha) = F cos (\theta)$
Now we can eliminate all of T1, T2 and T3 in terms of given propositions and finally ....
From 2nd equation
$T_2 sin (\alpha) = 0.002 kg wt.$
From 3rd equation,
$T_1 cos (\alpha) = 0.002 kg wt. - F sin (\theta)$
From 4th equation,
$T_1 sin (\alpha) = F cos (\theta)$
Hence,
$$tan (\alpha) = \frac {T_1 sin (\alpha)}
{T_1 cos (\alpha)}$$
$$ =\frac {F cos (\theta)}
{0.002 kg wt. - F sin (\theta)}$$
Have a wonderful day!
Best Answer
Yes, the tension in each string is $5g$ N because it has to support the particle's weight. The particle has that force pulling downward because of gravity, so has to have an upward force to balance it. The support is providing an upward force of $5g$ N to counteract the force at $R$. If the strings had mass the tension would be increasing as you go up because each bit of string would have to support the particle plus the weight of string below it.