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The time period of a simple pendulum of length L as measured in an elevator descending with acceleration $\frac{g}{3}$ is

$2 π \sqrt{\frac{3 L}{g}}$

$π \sqrt{(\frac{3 L}{g})}$

$2 π \sqrt{(\frac{3 L}{2 g})}$

$2 π \sqrt{\frac{2 L}{3 g}}$

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Explanation

(c) The effective acceleration in a lift descending with acceleration $\frac{g}{3} is g_{eff} = g - \frac{g}{3} = \frac{2 g}{3} so, T = 2 π \sqrt{(\frac{L}{g_{eff}})} = 2 π \sqrt{(\frac{L}{2 g / 3})} = 2 π \sqrt{(\frac{3 L}{2 g})}$

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If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period

$T = 2 π \sqrt{\frac{R_{e}}{g}}$

$T = 2 π \sqrt{\frac{2 R_{e}}{g}}$

$T = 2 π \sqrt{\frac{R_{e}}{2 g}}$

T=2 seconds

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Explanation

(a)

$A c c e l e r a t i o n d u e t o g r a v i t y a t a d e p t h d b e l o w t h e s u r f a c e o f t h e e a r t h; g' = g (1 - \frac{d}{R}) = \frac{g}{R} (R - d) = \frac{g}{R} x g' \propto x \Rightarrow ω^{2} = \frac{g}{R} T = 2 π \sqrt{\frac{R_{e}}{g}}$

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If the displacement equation of a particle be represented by $y = A \sin P T + B \cos P T$ , the particle executes

A uniform circular motion

A uniform elliptical motion

A S.H.M.

A rectilinear motion

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Explanation

$l e t A = r \cos θ, B = r \sin θ$

$\Rightarrow y = r \sin (P T + θ) w h i c h i s t h e e q u a t i o n o f S H M .$

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A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $Fsinωt$ . If the amplitude of the particle is maximum for $ω = ω_{1}$ and the energy of the particle is maximum for $ω = ω_{2}$ , then (where $ω_{0}$ is natural frequency of oscillation of particle)

$ω_{1} = ω_{0}$ and $ω_{2} \neq ω_{0}$

$ω_{1} = ω_{0}$ and $ω_{2} = ω_{0}$

$ω_{1} \neq ω_{0}$ and $ω_{2} = ω_{0}$

$ω_{1} \neq ω_{0}$ and $ω_{2} \neq ω_{0}$

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Explanation

Energy of particle is maximum at natural frequency i.e., $ω_{2} = ω_{0}$ .

For amplitude resonance (amplitude maximum)

$A = \frac{F_{o}}{\sqrt{m^{2} {(ω_{o}^{2} - ω_{d}^{2})}^{2} + {(b ω_{d})}^{2}}} F o r m a x i m u m A, \frac{d A}{d ω} = 0 S o l v i n g, ω_{o}^{2} - ω_{d}^{2} = \frac{b^{2}}{2 m^{2}} S o, ω_{1} < ω_{o} o r ω_{1} \neq ω_{o}$

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The displacement of a particle varies according to the relation $x = 4 (cosπ t + sinπ t) .$ The amplitude of the particle is

– 4

$4 \sqrt{2}$

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Explanation

(d) For given relation

Resultant amplitude= $\sqrt{4^{2} + 4^{2}}$ = $4 \sqrt{2}$

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A S.H.M. is represented by $x = 5 \sqrt{2} (\sin 2 πt + \cos 2 πt) .$ The amplitude of the S.H.M. is

10 cm

20 cm

$5 \sqrt{2}$ cm

50 cm

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Explanation

(a) $x = 5 \sqrt{2} (\sin 2 πt + \cos 2 πt) .$

$= 5 \sqrt{2} \sin 2 π t + 5 \sqrt{2} \cos 2 π t$

$x = 5 \sqrt{2} \sin 2 πt + 5 \sqrt{2} \sin (2 π t + \frac{π}{2})$

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Amplitude of a wave is represented by

$A = \frac{c}{a + b - c}$

Then resonance will occur when

$b = - c / 2$

b = 0 and a = c

C.
$b = - a / 2$
D.

None of these

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Explanation

(b) $A = \frac{c}{a + b - c}$ when ,b=0 , a=c

Amplitude $A \to \infty$ . This corresponds to resonance.

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The displacement of a particle varies with time as $x = 12 \sin w t - 16 \sin^{3} w t$ (in cm). If its motion is S.H.M., then its maximum acceleration is -

$12 ω^{2}$

$36 ω^{2}$

$144 ω^{2}$

$\sqrt{192} ω^{2}$

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Explanation

(b) $x = 12 \sin ω t - 16 \sin^{3} ω t$ = $4 [3 \sin ω t - 4 \sin^{3} ω t]$

= $4 [\sin 3 ω t] (b y u \sin g \sin 3 θ = 3 \sin θ - 4 \sin^{3} θ)$

$A c c e l e r a t i o n i s m a x i m u m w h e n x = A S o m a x i m u m a c c e l e r a t i o n : a_{m a x} = {(3 ω)}^{2} \times 4 = 36 ω^{2}$

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A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is $U (x) = k {[x]}^{3}$ , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is -

Proportional to $\frac{1}{\sqrt{a}}$

Independent of a

C.
Proportional to $\sqrt{a}$
D.

Proportional to $a^{3 / 2}$

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Explanation

(a)

$U = k {|x|}^{3} \Rightarrow F = - \frac{d U}{d x} = - 3 k {|x|}^{2} A c c e l e r a t i o n; a = - ω^{2} x \Rightarrow F = m a = \frac{d^{2} x}{d t^{2}} = - m ω^{2} x O n c o m p a r i n g : m ω^{2} = 3 k x \Rightarrow ω = \sqrt{\frac{3 k x}{m}} \Rightarrow T = \frac{2 π}{ω} = 2 π \sqrt{\frac{m}{3 kx}} Also, for SHM, x = asinωt \Rightarrow T = 2 π \sqrt{\frac{m}{3 k a s \sin ω t}}$

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The metallic bob of a simple pendulum has the relative density $ρ$ . The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is given by

$T \frac{ρ - 1}{ρ}$

$T \frac{ρ}{ρ - 1}$

$T \sqrt{\frac{ρ - 1}{ρ}}$

$T \sqrt{\frac{ρ}{ρ - 1}}$

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Explanation

(d) When the bob is immersed in water ,

its effective weight =

$(m g - \frac{m}{ρ} g) = m g (\frac{ρ - 1}{ρ}) s o g_{e f f} = g (\frac{ρ - 1}{ρ})$

$\frac{T}{T} = \sqrt{\frac{g}{g_{e f f}}} \Rightarrow T' = T \sqrt{\frac{ρ}{ρ - 1}}$

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Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.