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1

AIEEE 2003

MCQ (Single Correct Answer)
The length of a simple pendulum executing simple harmonic motion is increased by $$21\% $$. The percentage increase in the time period of the pendulum of increased length is
A
$$11\% $$
B
$$21\% $$
C
$$42\% $$
D
$$10\% $$

Explanation

$$T = 2\pi \sqrt {{\ell \over g}} $$ and $$T' = 2\pi \sqrt {{{1.21\ell } \over g}} $$

$$\left( \, \right.$$ as $$\ell ' = \ell + 21\% $$ of $$\left. \ell \right)$$

$$\% $$ increase $$ = {{T' - T} \over T} \times 100$$

$$ = {{\sqrt {1.21\ell } - \sqrt \ell } \over {\sqrt \ell }} \times 100$$

= $${{\sqrt {{{121} \over {100}}l} - \sqrt l } \over {\sqrt l }} \times 100$$

= $${{{{11} \over {10}}\sqrt l - \sqrt l } \over {\sqrt l }} \times 100$$

= $${{{{11} \over {10}} - 1} \over 1} \times 10$$

$$ = 10\% $$
2

AIEEE 2003

MCQ (Single Correct Answer)
Two particles $$A$$ and $$B$$ of equal masses are suspended from two massless springs of spring of spring constant $${k_1}$$ and $${k_2}$$, respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of $$A$$ and $$B$$ is
A
$$\sqrt {{{{k_1}} \over {{k_2}}}} $$
B
$${{{{k_2}} \over {{k_1}}}}$$
C
$$\sqrt {{{{k_2}} \over {{k_1}}}} $$
D
$${{{{k_1}} \over {{k_2}}}}$$

Explanation

Maximum velocity during $$SHM$$ $$ = A\omega = A\sqrt {{k \over m}} $$

$$\left[ {\,\,} \right.$$ $$\therefore$$ $$\omega = \sqrt {{k \over m}} $$ $$\left. {\,\,} \right]$$

Here the maximum velocity is same and $$m$$ is also same

$$\therefore$$ $${A_1}\sqrt {{k_1}} = {A_2}\sqrt {{k_2}} $$

$$\therefore$$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}} $$
3

AIEEE 2003

MCQ (Single Correct Answer)
A mass $$M$$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $$SHM$$ of time period $$T.$$ If the mass is increased by $$m.$$ the time period becomes $${{5T} \over 3}$$. Then the ratio of $${{m} \over M}$$ is
A
$${3 \over 5}$$
B
$${25 \over 9}$$
C
$${16 \over 9}$$
D
$${5 \over 3}$$

Explanation

$$T = 2\pi \sqrt {{M \over k}} $$

$$T' = 2\pi \sqrt {{{M + m} \over k}} = {{5T} \over 3}$$

$$\therefore$$ $$2\pi \sqrt {{{M + m} \over k}} $$

$$ = {5 \over 3} \times 2\pi \sqrt {{M \over k}} $$

$$ \Rightarrow {m \over M} = {{16} \over 9}$$
4

AIEEE 2002

MCQ (Single Correct Answer)
A child swinging on a swing in sitting position, stands up, then the time period of the swing will
A
increase
B
decrease
C
remains same
D
increases of the child is long and decreases if the child is short

Explanation

KEY CONCEPT : The time period $$T = 2\pi \sqrt {{\ell \over g}} $$ where

$$\ell $$ $$=$$ distance between the point of suspension and the center of mass of the child. This distance decreases when the child stands

$$\therefore$$ $$T' < T$$ i.e., the period decreases.

Questions Asked from Simple Harmonic Motion

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