A simple pendulum with length L and mass m of the bob is vibrating with an amplitude a. Then, the maximum tension in the string is

Question

A simple pendulum with length L and mass m of the bob is vibrating with an amplitude a. Then, the maximum tension in the string is (A)  2mkr2t  (B)  mkr2t2  (C)  mk2r2t  (D)  mk2rt2

Tardigrade Admin · Accepted Answer

Tension in the string T-mg cos θ =(mv2/L) T is maximum, when θ =0o ∴ T max =mg+(mv2 max /L)=mg+(ma2ω 2/L) ∵ a is amplitude T max =mg+(ma2/L)× (g/L) =mg[ 1+( (a/L) )2 ]

Q. A simple pendulum with length L and mass m of the bob is vibrating with an amplitude a. Then, the maximum tension in the string is

$2 mk r^{2} t$

$mk r^{2} t^{2}$

$m k^{2} r^{2} t$

$m k^{2} r t^{2}$

Tension in the string $T - m g cos θ = \frac{m v ^{2}}{L}$ T is maximum, when $θ = 0^{o}$
$∴$ $T_{m a x} = m g + \frac{m v ^{2} _{m a x}}{L} = m g + \frac{m a ^{2} ω ^{2}}{L}$ $∵$ a is amplitude $T_{m a x} = m g + \frac{m a ^{2}}{L} \times \frac{g}{L}$ $= m g [1 + (\frac{a}{L})^{2}]$

Q. A simple pendulum with length L and mass m of the bob is vibrating with an amplitude a. Then, the maximum tension in the string is

2mkr2t

mkr2t2

mk2r2t

mk2rt2

Tension in the string T−mgcosθ=Lmv2​ T is maximum, when θ=0o ∴ Tmax​=mg+Lmv2max​​=mg+Lma2ω2​ ∵ a is amplitude Tmax​=mg+Lma2​×Lg​ =mg[1+(La​)2]

$2 mk r^{2} t$

$mk r^{2} t^{2}$

$m k^{2} r^{2} t$

$m k^{2} r t^{2}$

Tension in the string $T - m g cos θ = \frac{m v ^{2}}{L}$ T is maximum, when $θ = 0^{o}$
$∴$ $T_{m a x} = m g + \frac{m v ^{2} _{m a x}}{L} = m g + \frac{m a ^{2} ω ^{2}}{L}$ $∵$ a is amplitude $T_{m a x} = m g + \frac{m a ^{2}}{L} \times \frac{g}{L}$ $= m g [1 + (\frac{a}{L})^{2}]$