Question

A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makes an angle of 37° with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.

Answer 1

given in the question,
Length of the string = 50 cm =0.5 m

Refer to the attachment
cosθ = xz /xy
xz = xycosθ
xz = (0.5 × cos37°)
xz = ( 0.5 × 0.8)
xz = 0.4
Now,
zn = 0.5-0.4
zn = 0.1 m

Energy on the n point = energy on the y point
$1/2 mv^2 = mg(zn)$ 
v² = 20 × 0.1
v² = 2


Now tension 
$T = \frac{mv^2}{r} +mg$ 
T = 0.1 [(2/0.5)  +10)]
T = 1.4 N.


Hope it Helps . :-)

Answer 2

The tension in the string when the bob is at its lowest position is 1.4 N.

Explanation:

Given data  in the question,

String’s length  = 50 cm        

Converting centimeter to meter

String’s length  $=\frac{50}{100}=0.5 \mathrm{m}$

                $\cos \theta=\frac{x z}{x y}$

                    xz = xycosθ

                    xz = (0.5 × cos37°)

                    xz = ( 0.5 × 0.8)

                    xz = 0.4

Now,

         $z_n$ = 0.5-0.4

         $z_n$ = 0.1 m

Energy in point n = energy  in the point y  

                    $\frac{1}{2} m v^{2}=m g(z n)$

                         $v^{2}=20 \times 0.1$

                         $v^{2}=2$

Now tension,

                    $\mathrm{T}=\frac{m v^{2}}{r}+m g$

                    $\left.\mathrm{T}=0.1\left[\left(\frac{2}{0.5}\right)+10\right)\right]$

                    $\mathrm{T}=0.1[(4+10)]$

                    T = 0.4 + 1

                    T = 1.4 N.

Therefore, the tension is 1.4 N.