[tex] \large \{ \overbrace {\underbrace{ \bold {\pmb { \: \: \blue{ QU ESTION } \: \: }}}} \}[/tex]


[tex] \large \{ \overbrace {\underbrace{ \bold {\pmb { \: \: \blue{ QU ESTION } \: \: }}}} \}[/tex]

Time Period of a Simple Pendulum in a Freely Falling Lift will be​

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Sophia

2 thoughts on “[tex] \large \{ \overbrace {\underbrace{ \bold {\pmb { \: \: \blue{ QU ESTION } \: \: }}}} \}[/tex]<br /><br /><br /><br”

  1. EXPLAINATION

    Formula for Time Period

    Time period, t = 2π √(t/g)

    Where

    • t time period
    • g acceleration due to gravity

    During free fall the body feels weightless inside the lift or no Acceleration will be acting on the body So, g = 0

    NOW

    ∴ Time period, t

    • = 2π √(t/0)
    • = 2π √∞
    • = ∞

    Thus, the time period of a simple pendulum in a freely falling lift will be infinity (∞).

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  2. [tex]{\huge{\boxed{\tt{\color {red}{Answer❀✿°᭄}}}}}[/tex]

    During free fall the body feels weightless inside the lift. Hence, the time period of a simple pendulum in a freely falling lift will be infite(∞).

    [tex]\huge\mathfrak\pink{itz \: Sujitha}[/tex]

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