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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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The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $θ$ , is given by -

$2 π \sqrt{\frac{L}{gcosθ}}$

$2 π \sqrt{\frac{L}{gsin θ}}$

$2 π \sqrt{\frac{L}{g}}$

$2 π \sqrt{\frac{L}{gtanθ}}$

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Explanation

When a simple pendulum is suspended from the roof of a vehicle moving down an inclined plane without friction, the effective acceleration due to gravity experienced by the pendulum is g cos(Î¸), where Î¸ is the angle of inclination. Therefore, the time period is T = 2Ï€âˆš(L/(g cos(Î¸))), where L is the length of the pendulum.

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One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire is A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to -

$2 π (\frac{m}{K})$

$2 π {\{\frac{(YA + KL) m}{YAK}\}}^{1 / 2}$

$2 π \frac{mYA}{KL}$

$2 π \frac{mL}{YA}$

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Explanation

b) The wire may be treated as a string for which force constant $k_{1} = \frac{F o r c e}{E x t e n c i o n} = \frac{Y A}{L} (Y = \frac{F}{A} \times \frac{L}{∆ L})$

Spring constant of the spring $k_{2} = K$

Hence spring constant of the combination (series)

$k_{e q} = \frac{k_{1} k_{2}}{k_{1} - k_{2}} = (\frac{(Y A / L) K}{(Y A / L) + K}) = \frac{Y A K}{Y A + K L}$

$⊝$ Time period $T = 2 π \sqrt{\frac{m}{k}} = 2 π {[\frac{(YA + KL) m}{YAK}]}^{1 / 2}$

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A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency $ω_{0}$ . An external force F (t) proportional to $\cos ω t (ω \neq ω_{0})$ is applied to the oscillator. The time displacement of the oscillator will be proportional to -

$\frac{m}{{ω^{2}}_{0} - ω^{2}}$

$\frac{1}{m ({ω^{2}}_{0} - ω^{2})}$

$\frac{1}{m ({ω^{2}}_{1} + ω^{2})}$

$\frac{m}{{ω^{2}}_{1} + ω^{2}}$

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Explanation

(b) For forced oscillation,

$x = x_{0} \sin (ω t + \emptyset) a n d F = F_{0} \cos ω t$

where, $x_{0} = \frac{F}{m ({ω^{2}}_{0} - ω^{2})} \propto \frac{1}{m ({ω^{2}}_{0} - ω^{2})}$

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The displacement y of a particle executing periodic motion is given by $y = 4 \cos^{2} (t / 2) \sin (1000 t) .$ This expression may be considered to be a result of the superposition of ........... independent harmonic motions

Two

Three

Four

Five

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Explanation

(b)

$y = 4 \cos^{2} (\frac{t}{2}) \sin 1000 t \Rightarrow y = 2 (1 + \cos t) \sin (1000 t) (A s 1 + \cos t = 1 + 2 \cos^{2} (\frac{t}{2}) - 1 = 2 \cos^{2} (\frac{t}{2})) \Rightarrow y = 2 \sin 1000 t + 2 \sin 1000 t \cos t \Rightarrow y = 2 \sin 1000 t + \sin (1000 t - t) + \sin (1000 t + t) [2 \sin A \cos B = \sin (A + B) + \sin (A - B)] \Rightarrow y = 2 \sin 1000 t + \sin 999 t + \sin 1001 t It is a sum of three S . H . M .$

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Equation of the wave is given by y = 0.4 sin(314t - 3.14x), where x and y are in meter and t is in second. The speed of the wave is :

50 m/s

100 m/s

314 m/s

3.14 m/s

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Explanation

$y = a \sin (ωt - kx) speed of wave = \frac{ω}{k} = \frac{314}{3.14} = 100 m / s$

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A plane progressive wave cannot be represented by

$y = a \sin (ωt \pm kx)$

$y = a \sin 2 π (\frac{t}{T} \mp \frac{x}{λ})$

$y = a \sin \frac{2 π}{λ} (Vt \mp x)$

$y = A \log x + B \log x$

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Explanation

$y = A \log x + B \log x i s n o t a w a v e .$

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The phase difference between the prongs of a tuning fork is :

$2 π$

$3 π$

$π$

$5 π$

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Explanation

2 $π$

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$The isothermal elasticity of a medium is E_{i} and the adiabatic elasticity in E_{a} . The velocity of sound in the medium is proportional to$

$E_{i}$

$E_{a}$

$\sqrt{E_{i}}$

$\sqrt{E_{a}}$

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Explanation

$v = \sqrt{\frac{E_{}}{ρ}} \{Laplace formula\} v \propto \sqrt{E_{a}}$

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When a wave is reflected from a denser medium the change in phase is :

$π$

$2 π$

$3 π$

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Explanation

When a wave undergoes a reflection at a denser medium then it's crest reflected as a trough and vice versa.

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