2 Marks Questions

Question 12 Marks

A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find, the length of a seconds pendulum at a place where $\text{g}=\pi^2\text{m/s}^2.$

Answer

$\text{T}=2\sec.$$\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$
$\Rightarrow2=2\pi\sqrt{\frac{\ell}{10}}$
$\Rightarrow\frac{\ell}{10}=\frac{1}{\pi^2}$
$\Rightarrow\ell=1\text{cm}$ $(\therefore\pi^2\approx10)$

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Question 22 Marks

It is proposed to move a particle in simple harmonic motion on a rough horizontal surface by applying an external force along the line of motion. Sketch the graph of the applied force against the position of the particle. Note that the applied force has two values for a given position depending on whether the particle is moving in positive or negative direction.

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Question 32 Marks

The angle made by the string of a simple pendulum with the vertical depends on time as $\theta=\frac{\pi}{90}\sin[(\pi\text{s}^{-1})\text{t}].$ Find the length of the pendulum if $\text{g}=\pi^2\text{m/s}^2.$

Answer

From the equation,$\theta=\pi\sin[\pi\sec^{-1}\text{t}]$
$\therefore\omega=\pi\sec^{-1}$ (comparing with the equation of SHM)
$\Rightarrow\frac{2\pi}{\text{T}}=\pi$
$\Rightarrow\text{T}=2\sec.$
We know that $\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$$\Rightarrow2=2\sqrt{\frac{\ell}{\text{g}}}$
$\Rightarrow1=\sqrt{\frac{\ell}{\text{g}}}$
$\Rightarrow\ell=1\text{m}.$
$\therefore$ Length of the pendulum is 1m.

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Question 42 Marks

A pendulum clock giving correct time at a place where $g = 9.800\ m/s^2$ is taken to another place where it loses $24$ seconds during $24$ hours. Find the value of $g$ at this new place.

Answer

For the pendulum, $\frac{\text{T}_1}{\text{T}_2}=\sqrt{\frac{\text{g}_2}{\text{g}_1}}$
Given that, $\text{T}_1=2\sec,\ \text{g}_1=9.8\ \text{m/s}^2$
$\text{T}_2=\frac{24\times3600}{\Big(\frac{24\times3600-24}{2}\Big)}$
$=2\times\frac{3600}{3599}$
Now, $\frac{\text{g}_2}{\text{g}_1}=\Big(\frac{\text{T}_1}{\text{T}_2}\Big)^2$
$\therefore\text{g}_2=(9.8)\Big(\frac{3599}{3600}\Big)^2$
$=9.795\ \text{m/s}^2$

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Question 52 Marks

A simple pendulum of length 40cm is taken inside a deep mine. Assume for the time being that the mine is 1600km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400km.

Answer

Length of the pendulum = 40cm = 0.4m. Let acceleration due to gravity be g at the depth of 1600km.$\therefore\text{gd}=\text{g}(1-\frac{\text{d}}{\text{R}})=9.8\Big(1-\frac{1600}{6400}\Big)$
$=9.8\Big(1-\frac{1}{4}\Big)=9.8\times\frac{3}{4}=7.35\text{m/s}^2$
$\therefore$ Time period $\text{T}'=2\pi\sqrt{\frac{\ell}{\text{g}\delta}}$
$=2\pi\sqrt{\frac{0.4}{7.35}}=2\pi\sqrt{0.054}=2\pi\times0.23$
$=2\times3.14\times0.23=1.465\approx1.47\sec.$

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Question 62 Marks

A particle executes simple harmonic motion with an amplitude of 10cm. At what distance from the mean position are the kinetic and potential energies equal?

Answer

r = 10cm Because, K.E. = P.E. So, $\Big(\frac{1}{2}\Big)\text{m}\omega^2(\text{r}^2-\text{y}^2)=\Big(\frac{1}{2}\Big)\text{m}\ \omega^2\text{y}^2$$\text{r}^2-\text{y}^2=\text{y}^2$
$\Rightarrow2\text{y}^2=\text{r}^2$
$\Rightarrow\text{y}=\frac{\text{r}}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}\text{cm}$ From the the mean position.

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Question 72 Marks

A hollow sphere filled with water is used as the bob of a pendulum. Assume that the equation for simple pendulum is valid with the distance between the point of suspension and centre of mass of the bob acting as the effective length of the pendulum. If water slowly leaks out of the bob, how will the time period vary?

Answer

first it will increase then it will decrease.

$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ first l increases and then it decreases.

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Question 82 Marks

A particle executes simple harmonic motion. If you are told that its velocity at this instant is zero, can you say what is its displacement? If you are told that its velocity at this instant is maximum, can you say what is its displacement?

Answer

No, to know the displacement we should know the initial position of the particle. If it is said that initial position is mean position then we can answer the above question.

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Question 92 Marks

The force acting on a particle moving along $X-$axis is $F = -k(x - u_0t)$ where $k$ is a positive constant. An observer moving at a constant velocity $v_0,$ along the $X-$axis looks at the particle. What kind of motion does he find for the particle?

Answer

since he moves with constant velocity, he sees the same force. And this force is not that of $\text{SHM}.$

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Question 102 Marks

In measuring time period of a pendulum, it is advised to measure the time between consecutive passage through the mean position in the same direction. This is said to result in better accuracy than measuring time between consecutive passage through an extreme position. Explain.

Answer

Because mean position is fixed, while extreme position keeps on changing. So, when we use stop watch, we are very sure about the mean position.

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Question 112 Marks

A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum i.e., a pendulum having frequency same as that of the block.

Answer

Time period of simple pendulum $=2\pi\sqrt{\frac{1}{\text{g}}}$ Time period of spring is $2\pi\sqrt{\frac{\text{m}}{\text{k}}} T_p = T_s [$Frequency is same$]$
$\Rightarrow\sqrt{\frac{1}{\text{g}}}=\sqrt{\frac{\text{m}}{\text{k}}}$
$\Rightarrow\frac{1}{\text{g}}=\frac{\text{m}}{\text{k}}$
$\Rightarrow1=\frac{\text{mg}}{\text{k}}=\frac{\text{F}}{\text{k}}=\text{x}. ($Because, restoring force $=$ weight $= F = mg)$
$\Rightarrow1=\text{x }(\text{proved})$

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Question 122 Marks

The pendulum of a clock is replaced by a spring-mass system with the spring having spring constant 0.1N/m. What mass should be attached to the spring?

Answer

$\text{k}=0.1\text{N/m}$$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\sec$ [Time period of pendulum of a clock = 2 sec]
So, $4\pi^{2+}\Big(\frac{\text{m}}{\text{k}}\Big)=4$
$\therefore\text{m}=\frac{\text{k}}{\pi^2}=\frac{0.1}{10}=0.01\text{kg}\approx10\text{gm}.$

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