Oscillations NEET प्रश्न हिंदी में — भौतिक विज्ञान (Physics) MCQ उत्तर सहित

$K_1 l _ 1 = K_2 l_2 = kl $ $ \therefore K_1 \left( l \over 4 \right) = K_2 \left( { 3 \over 4 } l \right) = kl $ $ force constant of spring having length {3 \over 4 } l \;in $ $ k_2 = { 4 \over 3 } k $

$ T \alpha \sqrt 1 ^1 $ $ because \,2 \pi and\, g are \,constants $ $ \therefore { T_2 \over T_1} = \sqrt { l_2 \over l_1 } = \sqrt {1.2 l_1 \over l_1 } =1.1 $ $ \therefore \% increase = { T_2 -T_1 \over T_1 } \times 100 = 10 \% $

$ y = A sin ( \omega t + \phi ) $ $ { A \over \sqrt 2 } = A sin \omega t \{ \phi = 0 \} \Rightarrow { 1 \over \sqrt 2 } = sin \omega t = { \pi \over 4 } $ $ \therefore { 2 \pi \over T } .t = { \pi \over 4 } $ $ \therefore t = { T \over 8 } $

$ T_1 = 2 \pi \sqrt { m^1 \over k_1 } \Rightarrow k_1 = { 4 \pi^2 M \over T_1^2 } and k_2 = { 4 \pi ^2 M \over T_2^2 } $ श्रेणी संयोजन के लिए ; $ T = 2 \pi \sqrt { M \over k} where k = { k_1 k_2 \over k_1 + k_2 } $ $ \therefore T = 2 \pi \sqrt { {M^1 \over 4 \pi^2 M} ( T_1^2 + T_2 ^2 ) } $ $ \therefore T = \sqrt { T_1^2 + T_2 ^2 } \Rightarrow T^2 = T_1 ^2 + T_2 ^2 $

$ \nu_1 = { dy_1 \over dt } $ $ and \nu_2 = { dy_2 \over dt } $

$ here y = kt^2 $ $ \therefore { dy \over dt } = 2 kt \Rightarrow { d^2 y \over dt^2 } = 2k = 2 ms^{-2} $ $ \therefore the point of support in \;moving\; upwards with\; an acceleration\; of 2 m/s^2 $ $ \therefore effective acceleration g' = g + a = 12 m/s^2 $ $ now \;T_1 = 2 \pi \sqrt { l \over g } and\; T_2 = 2 \pi \sqrt { l \over g } $

$ \nu = w \sqrt { A^2 -x^2 } $ x=o से x = A /2 तक चलने के लिए वेग x =A /2 से x = A की तुलना में अधिक होगा $ \therefore T_1 \lt T_2 $

$ Angular frequency \omega = \sqrt { k \over m} $ $ Since 'm' is constant , \omega \alpha \sqrt k ^1 $ $Now , a _{max} = A \omega^2 \Rightarrow \omega = \sqrt { a_{max} \over A } $ $ \therefore {a_{max} \over A} = k \Rightarrow { a_{max} \over K } = A $ $ \therefore A \alpha {1 \over k } $

$ For a spring , T = 2 \pi \sqrt { m \over k } \Rightarrow T \alpha { 1 \over \sqrt k } $ ( m नियत है )

$ Phase of 1st oscillator \theta_1 = \omega _1 t + \phi = { 2 \phi \over T_1 } t + \phi $ $ For 2nd oscillator , \theta_2 = \omega_2 t + \phi = {2 \phi \over T_2 } t + \phi $ $ Phase diff \theta_1 - \theta_2 $