The mass of the bob of a simple pendulum of length L is m . If the bob is left from its horizontal position then the speed of the bob and the tension in the thread in the lowest position of the bob will be respectively.

To find the speed of the bob and the tension in the thread at the lowest position, we can apply the principles of conservation of mechanical energy and circular motion dynamics.

Step 1: Find the speed of the bob at the lowest position
Let the lowest position of the bob be the reference level for potential energy, where potential energy is zero ($U_{\text{lowest}}=0$).
Initially, the bob is released from the horizontal position. At this position, it is at a height equal to the length of the pendulum $L$ above the lowest point. Therefore, its initial gravitational potential energy is:
$U_{\text{initial}}=mgL$
Since the bob is released from rest, its initial kinetic energy is:
$K_{\text{initial}}=0$
At the lowest position, let the speed of the bob be $v$. The kinetic energy is:
$K_{\text{lowest}}=\frac{1}{2}m{v}^2$

By the law of conservation of mechanical energy:
$E_{\text{initial}}=E_{\text{lowest}}$
$mgL+0=0+\frac{1}{2}m{v}^2$
Simplifying the equation by dividing both sides by $m$:
$gL=\frac{1}{2}$
$v^2=2gL$
$v=\sqrt{2gL}$

Step 2: Find the tension in the thread at the lowest position
At the lowest position, two forces act on the bob vertically:
1. The tension $T$ in the thread acting vertically upwards.
2. The gravitational force (weight) $mg$ acting vertically downwards.

Since the bob is moving in a circular path of radius $L$, the net force towards the center (upwards) provides the necessary centripetal force:
$T-mg=\frac{m{v}^2}{L}$

Substitute the value of $v^2=2gL$ into the equation:
$T-mg=\frac{m\left(2gL\right)}{L}$
$T-mg=2mg$
$T=2mg+mg$
$T=3mg$

Therefore, the speed of the bob is $\sqrt{2gL}$ and the tension in the thread is $3mg$.