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2 months ago

Physics Oscillations Class 11th

If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is

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alok kumar singh

Contributor-Level 10

$\frac{Δ T}{T} = \frac{1}{2} \frac{Δ l}{l}$

$\Rightarrow Δ T = \frac{1}{2} * \frac{0.1}{100} * 24 * 60 * 60 = 43.2 s$

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Physics Oscillations Oscillations

The displacement simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is:

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Vishal Baghel

Contributor-Level 10

$x = A s i n ω t$

Angle travelled by while moving from A to B

$θ = 30 ° = \frac{π °}{6}$

$θ = ω t$

$\frac{π}{6} = ω * 3 [t = 3 s e c]$

$ω = \frac{π}{18}$

$\Rightarrow T = \frac{2 π}{ω} = \frac{2 π * 18}{π} = 36 s e c$

T = 36 sec

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2 months ago

Physics Oscillations Oscillations

In figure (A), mass '2 m' is fixed on mass 'm' which is attached to two spring of spring constant k. In figure (B), mass 'm' is attached to two springs of spring constant 'k' and '2k'. If mass 'm' in (A) and in (B) are displaced by distance 'x' horizontally and then released, then time period T₁ and T₂ corresponding to (A) and (B) respectively follow the relation.

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Vishal Baghel

Contributor-Level 10

$T_{1} = 2 π \sqrt[]{\frac{3 m}{2 k}}$

$T_{2} = 2 π \sqrt[]{\frac{m}{3 k}}$

$\frac{T_{1}}{T_{2}} = \frac{2 π \sqrt[]{\frac{3 m}{2 k}}}{2 π \sqrt[]{\frac{m}{3 k}}} = \frac{3}{\sqrt[]{2}}$

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2 months ago

Physics Oscillations Class 11th

The length of a seconds pendulum at a height h = 2R from earth surface will be:

(Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $π^{2} m s^{- 2}$ )

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Payal Gupta

Contributor-Level 10

As we know that

$g_{P} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{R + 2 R})}^{2} = \frac{g}{9}, s o$

$T = 2 π \sqrt[]{\frac{l}{g_{P}}} \Rightarrow l = \frac{g_{p} T^{2}}{4 π^{2}} = \frac{1}{9} m$

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2 months ago

Physics Oscillations Class 11th

The metallic bob of simple pendulum has the relative density. 5 The time period of this pendulum is 10 s. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt[]{x} s .$ The value of x will be_______.

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alok kumar singh

Contributor-Level 10

g_eff = g – $(\frac{ρ_{w}}{ρ_{b}}) g$

$T = 2 π \sqrt[]{\frac{l}{g}}$

$T' = 2 π \sqrt[]{\frac{l}{g_{e f f}}}$

$\Rightarrow T' = \sqrt[]{\frac{5}{4}} T$

= $\sqrt[]{\frac{5}{4}} * 10$

= $5 \sqrt[]{5} s e c$

So, x = 5

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2 months ago

Physics Oscillations Class 11th

A mass 0.9kg, attached to a horizontal spring, executes SMH with an amplitude A₁. When this mass passes through its mean position, then a smaller mass of 124 g is placed over it both masses move together with amplitude A₂. If the ratio $\frac{A_{1}}{A_{2}} i s \frac{α}{α - 1},$ then the value of a will be______.

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Payal Gupta

Contributor-Level 10

$m_{1} = 900 g m, w_{1} = \sqrt[]{\frac{k}{. 9}}$

$m_{2} = 1024 g m, ω_{2} = \sqrt[]{\frac{k}{1.024}}$

$m_{1} v_{1} = m_{2} V_{2}$

$\Rightarrow m_{1} ω_{1} A_{1} = m_{2} ω_{2} A_{2}$

$\Rightarrow \frac{A_{1}}{A_{2}} = \sqrt[]{\frac{1024}{900}} = \frac{32}{30} = \frac{16}{15} = \frac{16}{16 - 1}$

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2 months ago

Physics Oscillations Class 11th

The speed of a transverse wave passing through a string of length 60 cm and mass 10g is 60 ms^-1. The area of cross-section of the wire is 2.0 mm² and its Young's modulus is 1.2 * 10¹¹ Nm^-2. The extension of the wire over its natural length due to its tension will be x * 10^-5 m. The value of x is________.

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Payal Gupta

Contributor-Level 10

m = 10g

$l = 50 c m$

A = 2mm²

Y = 1.2 * 10¹¹N/m²

$Δ x = x * 1 0^{- 5} m$

As, $\frac{T}{A} = Y \frac{Δ x}{l}$

$Δ x = \frac{T l}{A Y}$

$T l = V^{2} m$

= $\frac{V^{2} m}{A Y}$

$= \frac{3600 * 10 * 1 0^{- 3}}{2 * 1 0^{- 6} * 1.2 * 1 0^{11}} = \frac{1800 * 1 0^{- 3} * 1 0^{6}}{1.2}$

$= \frac{18}{12} * 1 0^{- 4} = \frac{3}{2} * 1 0^{- 4} = 15 * 1 0^{- 5} m$

So, x = 15

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2 months ago

Physics Oscillations Class 11th

Two massless springs with spring constants 2k and 9k, carry 50g and 100g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then the ratio of their respective amplitudes will

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Payal Gupta

Contributor-Level 10

$ω_{1} = \sqrt[]{\frac{k}{m}} = \sqrt[]{\frac{2 k}{0.05}}$

$\begin{array}{l} \Rightarrow A_{1} ω_{1} = A_{2} ω_{2} \\ \frac{A_{1}}{A_{2}} = \frac{ω_{2}}{ω_{1}} = \sqrt[]{\frac{9 k}{0.1}} * \sqrt[]{\frac{0.05}{2 k}} \end{array}$

$\frac{A_{1}}{A_{2}} = \frac{3}{2}$

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2 months ago

Physics Oscillations Class 11th

The motion of a simple pendulum executing S.H.M. is represented by the following equation. $y = A s i n (π t + ?),$ where time is measured in second. The length of pendulum is

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Payal Gupta

Contributor-Level 10

According to question, we can write

$ω = π = \sqrt[]{\frac{g}{l}} \Rightarrow l = \frac{g}{π^{2}} = \frac{9.8}{{(3.14)}^{2}} = 0.99395 = 99.4 c m$

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2 months ago

Physics Oscillations Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

A body is performing simple harmonic with an amplitude of 10cm. The velocity if the body was tripled by air Jet when it is at 5cm from its mean position. The new amplitude vibration is $\sqrt[]{x} c m .$ The value of x is__________.

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alok kumar singh

Contributor-Level 10

v = $ω \sqrt[]{A^{2} - x^{2}}$ $^{}^{}$

at x = 5, A = 10

$v' = 3 v = 3 ω \sqrt[]{A^{2} - 5^{2}} = ω \sqrt[]{A'^{2} - 5^{2}}$

$= 3 \sqrt[]{A^{2} - 5^{2}} = \sqrt[]{A'^{2} - 5^{2}}$

$\sqrt[]{1 0^{2} - 5^{2}} = \sqrt[]{A'^{2} - 25}$

$A'^{2} = 25 + 9 * 75$ $A'^{2} = 700$

$A' = \sqrt[]{700} c m$

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