JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 20)
A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is :
$${1 \over 4}Hz$$
$${1 \over {2\sqrt 2 }}Hz$$
$${1 \over 2}Hz$$
$$2$$ $$Hz$$
Explanation
For 1 kg block :
_8th_April_Morning_Slot_en_20_1.png)
Here frequency of spring (f) = $${1 \over {2\pi }}\sqrt {{k \over m}} $$
Given that, F = 1 Hz
$$\therefore\,\,\,$$ $${1 \over {2\pi }}\sqrt {{k \over 1}} $$ = 1
$$ \Rightarrow $$$$\,\,\,$$ k = 4$$\pi $$2 N m$$-$$1
For 8 kg Block :
_8th_April_Morning_Slot_en_20_2.png)
Here two identical springs are attached in parallel. So,
Keq = k + k = 2k
$$\therefore\,\,\,$$ Frequency of 8 kg block,
F' = $${1 \over {2\pi }}\sqrt {{{{k_{eq}}} \over {m'}}} $$
= $${1 \over {2\pi }}\sqrt {{{2k} \over 8}} $$
= $${1 \over {2\pi }}\sqrt {{{2 \times 4{\pi ^2}} \over 8}} $$
= $${1 \over {2\pi }} \times \pi $$
= $${1 \over 2}$$ Hz
Here frequency of spring (f) = $${1 \over {2\pi }}\sqrt {{k \over m}} $$
Given that, F = 1 Hz
$$\therefore\,\,\,$$ $${1 \over {2\pi }}\sqrt {{k \over 1}} $$ = 1
$$ \Rightarrow $$$$\,\,\,$$ k = 4$$\pi $$2 N m$$-$$1
For 8 kg Block :
Here two identical springs are attached in parallel. So,
Keq = k + k = 2k
$$\therefore\,\,\,$$ Frequency of 8 kg block,
F' = $${1 \over {2\pi }}\sqrt {{{{k_{eq}}} \over {m'}}} $$
= $${1 \over {2\pi }}\sqrt {{{2k} \over 8}} $$
= $${1 \over {2\pi }}\sqrt {{{2 \times 4{\pi ^2}} \over 8}} $$
= $${1 \over {2\pi }} \times \pi $$
= $${1 \over 2}$$ Hz
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