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11th Standard English Medium Physics Subject Oscillations Book Back 1 Mark Questions with Solution Part - I updated Book back Questions

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Std: 11th Standard - Physics

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Questions:

Multiple Choice Question

5 x 1 = 5

A mass of 3 kg is attached at the end of a spring moves with simple harmonic motion on a horizontal frictionless table with time period 2π and with amplitude of 2m, then the maximum fore exerted on the spring is

(a)

1.5 N

(b)

3 N

(c)

6 N

(d)

12 N

When a damped harmonic oscillator completes 100 oscillations, its amplitude is reduced to \(\frac{1}{3}\) of its initial value. What will be its amplitude when it completes 200 oscillations?

(a)

\(\frac{1}{5}\)

(b)

\(\frac{2}{3}\)

(c)

\(\frac{1}{6}\)

(d)

\(\frac{1}{9}\)

A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom, the period of oscillation will

(a)

first increase and then decrease

(b)

first decrease and then increase

(c)

increase continuously

(d)

decrease continuously

An ideal spring of spring constant k, is suspended from the ceiling of a room and a block of mass M is fastened to its lower end. If the block is released when the spring is un-stretched, then the maximum extension in the spring is

(a)

4\(\frac { Mg }{ k } \)

(b)

\(\frac { Mg }{ k } \)

(c)

2\(\frac { Mg }{ k } \)

(d)

\(\frac { Mg }{ 2k } \)

A spring is connected to a mass m suspended from it and its time period for vertical oscillation is T. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is

(a)

T'= \(\sqrt{2}\)T

(b)

\(T'=\frac { T }{ \sqrt { 2 } } \)

(c)

T'=\(\sqrt{2T}\)

(d)

\(T'=\sqrt { \frac { T }{ 2 } } \)

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Answers Key:

Multiple Choice Question

5 x 1 = 5

Amplitude at any instant is given by
\(a=a_{0} e^{-b t}\)
\(\mathrm{a}_{0}=\text { initial amplitude }\)
b = damping constant
Case 1:
\(t =100 \mathrm{~T} \)
\(a =\frac{a_{0}}{3} \)
\(\therefore \frac{a_{0}}{3} =a_{0} e^{-b 100 T} \)
\(e^{-b 100 T}=\frac{1}{3}\) ....(1)
Case 2:
\(a=a_{0} e^{-200 b T} \)
\(a=a_{0}^{(-100 b T) 2} e=a_{0}\left(\frac{1}{3}\right)^{2} \)
\(a=\frac{a_{0}}{9} \)

Initially when the sphere in completely filled with water, its centre of gravity (C.G) lies at its centre. As water flows out, the centre of gravity begins to shift below the centre of the sphere. The effective length of the pendulum increases and hence the time period increases. When the sphere is half empty C.G begins to rise up. As the length of pendulum decreases T decreases.

\(\mathrm{F} =-\mathrm{kx} \)
\(\therefore \mathrm{x} =\left|-\frac{F}{k^{\prime}}\right|=\frac{F}{k^{\prime}} \)
\(\mathrm{F} =\mathrm{Mg} \text { and } k^{\prime}=\frac{k}{2} \)
\(\therefore \mathrm{x} =\frac{M g}{\frac{k}{2}} \)
\(=\frac{2 M g}{k} \)

\(\mathrm{T}=2 \pi \sqrt{\frac{m}{k}}\)
when the spring is cut into two equal halves, than the force constant of each part is 2k. When the mass is suspended from one of the halves, new time period is \(T^{\prime}=2 \pi \sqrt{\frac{m}{2 k}}=\frac{T}{\sqrt{2}}\)