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A spring having length l and spring constant k is divided into two parts having lengths $l_1$ and $l_2$. If $l_1 = nl_2$, the force constant of the spring having length $l_2$ is
$ k_2 be the spring constant of the spring having length l_2$ $ Now , l_1 + l_2 = l $ $ n l_2 + l_2 = l $
When a mass m is suspended from the free end of a massless spring having force constant k, its oscillates with frequency f. Now if the spring is divided into two equal parts and a mass 2m is suspended from the end of anyone of them, it will oscillate with a frequency equal to ………….
$ f = { 1 \over 2 \pi } \sqrt { k \over m } and f' = { 1 \over 2 \pi } \sqrt { 2k \over 2m } $ $ { k' = 2 k } $ $ \therefore f' = f $
A body of mass 1 kg suspended from the free end of a spring having force constant $400 Nm^{-1}$ is executing S.H.M. When the total energy of the system is 2 joule, the maximum acceleration is
………$ms^{ – 2}$ .
Energy stoved =Work done $ \therefore E = { 1 \over 2 } k A^2 $ Now maximum acceleration $ a_{max} = \omega^2 A $
A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass 2 kg is attached. The length of the spring is 35 cm. Now when the turn table is rotated with an angular speed of $10 rad s^{– 1}$ , the length of the spring becomes 40 cm then the force constant of the spring is..... N/m.
Radius of the rotational motion r =0.4 m When the turn table rotates, the restoring force developed in the spring = centrifugal force $ \therefore F_{restore} = m \omega^2 r = 2 ( 10 ) ^2 \times 0.4 = 80 N $ Now increase in length of spring = 40-35 = 5 cm $ \therefore Force constant k - { F \over x } = { 80 \over 0.05 } = 1.6 \times 10^3 N/m $
A simple pendulum is executing S.H.M. around point O between the end points B and C with a periodic time of 6 s. If the distance between B and C is 20 cm then in what time will the bob move from C to D? Point D is at the mid-point of C and O.
Here T=6 s Amplitude OB = OC = 1/2 BC = 10cm $ \therefore OD =5 cm$ $ Now displacement x = A sin ( wt + \phi) ...(1) $ $ where A = 10 gm , \omega = { 2 \pi \over T } = { \pi \over 3 } rad $ Now if at t = 0 , oscillator is at C i.e at t =0 , x = A $\therefore A= Asin(\omega \times 0 + \phi) (OR) A=Asin \phi$ $ \Rightarrow sin \phi = 1 $ $ \Rightarrow \phi ={\pi \over 2} $ putting this in eqn (1) $ x = A sin ( \omega t + { \pi \over 2 } ) = A cos \omega t = 10 cos \omega t $ $ \therefore for x = 5 cm $ $ 5 = 10 cos \omega t \Rightarrow cos \omega t = { 1 \over 2 } $ $ \therefore \omega t = { \pi \over 3 } $ $ \therefore t = 1 S $
A small spherical steel ball is placed at a distance slightly away from the center of a concave mirror having radius of curvature 250 cm. If the ball is released, it will now move on the curved surface. What will be the periodic time of this motion? Ignore frictional force and take $g = 10 m / s^2$
Force responsible for oscillation in $ F = mg sin \theta = mg \theta $ $ \{ \theta is small \} $ $ = mg .{ x \over R} $ Comparing this with $ F = - kx $ $ k = { mg \over R } $
A simple pendulum having length l issus pended at the roof of a train moving with constant acceleration‘a’ along horizontal direction. The periodic time of this pendulum
Here 2 acceleration vectors g. and a are acting along mutually prependicular direction . $ \therefore effective acceleratioin l^n g_{eff} = \sqrt { g^2 + a^2 } $ $ \therefore T = 2 \pi \sqrt { l \over g _{eff} } $
A trolley is sliding down a frictionless slope having inclination è. If a simple pendulum is suspended on top of this trolley, its periodic time is given by
$ g^2 _{off} = a_x^2 + (g -ay ) ^2 here a_x g sin \theta cos \theta , ay = g sin ^2 \theta $ $ = a_x^2 + g^2 + a_y^2 - 2 ga_y $ $ = a^2 sin ^ 2 \theta cos ^2 \theta + g^2 + g^2 sin^2 \theta - 2 g^2 sin ^2 \theta $ $ = g^2 ( 1 -sin^2 \theta ) $ $= g ^2 cos ^2 \theta $ $ \therefore g_{eff} = g cos \theta $
A system is executing S.H.M. The potential energy of the systemfor displacement x is $E_1$ and for a displacement of y, the potential energy of the system is $E_2$. The potential energy for a displacement of (x+y) is ………
$ E_1 = {1 \over 2 } m \omega^2 x^2 \Rightarrow \sqrt E_1 = x \sqrt { {1 \over 2 } m \omega^2 } ...(1) $ $ E_2 = {1 \over 2 } m \omega^2 y^2 \Rightarrow \sqrt E_2 = x \sqrt { {1 \over 2 } m \omega^2 } ...(2) $ $ E = {1 \over 2 } m \omega^2 (x + y )^2 \Rightarrow \sqrt E = (x + y ) \sqrt { {1 \over 2 } m \omega^2 } ...(3) $ From (1) ,(2) ,(3) , $ \sqrt E = \sqrt E_1 + \sqrt E_2 $ $or E = E_1 + E_2 + 2 \sqrt { E_1 E_2 } $
A system is executing S.H.M. with a periodic time of 4/5 s under the influence of force $F_1$. When a force $F_2$ is applied, the periodic time is (2/5) s. Now if $F_1$ and $F_2$ are applied simultaneously along the same direction, the periodic time will be ………
$ \omega_1^2 = { k \over m } = { kx \over mx } = { F_1 \over mx} ...(1) $ $ similarly , \omega_2^2 = { F_2 \over mx} ...(2) $ $ if F_1 and F_2 acts simultaneously ,then angular frequency $ $ w_2 ={ F_1 + F_2 \over mx } .....(3) $ $ From (1) , (2) and (3) ; \omega^2 = \omega_1^2 + \omega_2^2 $ $ now ,use eqn \omega = { 2 \pi \over T} $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.