Question

A simple pendulum with length $\phi$ and bob of mass m executes SHM of small amplitude A. The maximum tension in the string will be:

• A. $mg(1+A/\phi )$
• B. $mg(1+A/\phi {)}^2$
• C. $mg(1+A^2/{\phi }^2)$
• D. 2 mg

Answer 1

The correct option is C $mg(1+A^2/{\phi }^2)$

B is the mean position ,'C' displace position.The amplitude 'A'.

From fig(2) A = CD, OC=l

If mass of the pendulum has a velocity 'v'(linear)then centrifugal force and component '$mg\cos \theta$' will balance the tension of string

$T=mg\cos \theta +\frac{m{v}^2}{l}$

Now T will be max when $\cos \theta =1^{\prime }$

$T_{max}=mg+\frac{m{v}^2}{l}$

(it is the mean position as$\cos \theta$ = 1 mean $\theta =0$

If 'as' is angular frequency

$\omega =\omega =\sqrt{g/l}$

and v is equal to '$\omega (a^2-x^2{)}^{\frac{1}{2}}$'. 'x' displacement of particle.

So, at mean position

$v=\omega a$

$\Rightarrow v=a\sqrt{g/l}\Rightarrow v^2=\frac{a^{2}g}{l}$

$T_{max}=mg+\frac{mg{a}^2}{l^2}=mg(1+{\left(\frac{a}{l}\right)}^2)$