2 Marks Questions

Question 512 Marks

Answer the following questions:Time period of a particle in SHM depends on the force constant k and mass m of the particle: $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}.$ A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

Answer

The time period of a simple pendulum, $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$For a simple pendulum, k is expressed in terms of mass m, as:
$\text{k}\propto\text{m}$
$\frac{\text{m}}{\text{k}}=\text{Constant}$
Hence, the time period T, of a simple pendulum is independent of the mass of the bob.

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Question 522 Marks

Why can't we use a pendulum to work as a clock in a satellite?

Answer

Time period of an oscillating pendulum changes with acceleration due to gravity. It is zero in a satellite. So, only clocks with spring can be used.

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Question 532 Marks

A vertical U-tube of uniform cross-section contains water upto a height of 20cm. Calculate the time period of the oscillation of water when it is disturbed.

Answer

The length of the liquid column, L = 2 × 20 = 40cm Time period of oscillation,$\text{T}=2\pi\sqrt{\frac{\text{L}}{2\text{g}}}=2\pi\sqrt{\frac{40}{2\times980}}=\frac{2\pi}{7}$
$=0.9\text{s}$

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Question 542 Marks

Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion ($\omega$ is any positive constant):$\sin^3\omega\text{t}$

Answer

Periodic, but not SHM The given function is:$\sin^3\omega\text{t}$
$=\frac{1}{2}[3\sin\omega\text{t}-\sin3\omega\text{t}]$
The terms $\sin\omega\text{t}$ and $\sin\omega\text{t}$ individually represent simple harmonic motion (SHM).
However, the superposition of two SHM is periodic and not simple harmonic.

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Question 552 Marks

The maximum acceleration of a simple harmonic oscillator is $\text{a}_0$ and the maximum velocity is $\text{v}_0$. What is the displacement amplitude?

Answer

Let A be the displacement amplitude and $\omega$ be the angular frequency of the simple harmonic oscillator. Then, $\text{a}_0=\omega^2\text{A}$$\text{v}_0=\omega\text{A}$
On dividing $\frac{\text{v}_0^2}{\text{a}_0}=\frac{\omega^2\text{A}^2}{\omega^2\text{A}}=\text{A}$$\text{A}=\frac{\text{v}_0^2}{\text{a}_0}$

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Question 562 Marks

What is the length of a simple pendulum, which ticks seconds ?

Answer

From Eq. (13.26), the time period of a simple pendulum is given by,
$
T=2 \pi \sqrt{\frac{L}{g}}
$

From this relation one gets,
$
L=\frac{g T^2}{4 \pi^2}
$

The time period of a simple pendulum, which ticks seconds, is $2 s$. Therefore, for $g=9.8 m s ^{-2}$
$
\text { and } \begin{aligned}
T & =2 s , L \text { is } \\
& =\frac{9.8\left( m s ^{-2}\right) \times 4\left( s ^2\right)}{4 \pi^2} \\
& =1 m
\end{aligned}
$

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Question 572 Marks

Which of the following functions of time represent $(a)$ simple harmonic motion and $(b)$ periodic but not simple harmonic? Give the period for each case.
$(1) \sin \omega t-\cos \omega t$
$(2) \sin ^2 \omega t$

Answer

$(a)\sin \omega t-\cos \omega t$
$= \sin \omega t-\sin (\pi / 2-\omega t)$
$= 2 \cos (\pi / 4) \sin (\omega t-\pi / 4)$
$= \sqrt{2} \sin (\omega t-\pi / 4)$
This function represents a simple harmonic motion having a period $T=2 \pi / \omega$ and a phase angle $(-\pi / 4)$ or $(7 \pi / 4)$
$(b)\sin ^2 \omega t$
$=1 / 2-1 / 2 \cos 2 \omega t$
The function is periodic having a period $T=\pi / \omega$.
It also represents a harmonic motion with the point of equilibrium occurring at $1 / 2$ instead of zero.

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Physics 2 Marks Questions