Question 141: Oscillations

Question

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 :

Accepted Answer

For 1 kg  block :
Here frequency of spring (f) = $${1 \over {2\pi }}\sqrt {{k \over m}} $$
Given that, F = 1 Hz
$$	herefore\,\,\,$$  $${1 \over {2\pi }}\sqrt {{k \over 1}} $$ = 1
$$ \Rightarrow $$$$\,\,\,$$ k = 4$$\pi $$
2
N m
$$-$$1
For 8 kg Block :
Here two identical springs are attached in parallel. So,
K
eq
= k + k = 2k
$$	herefore\,\,\,$$ Frequency of 8 kg block,
F' = $${1 \over {2\pi }}\sqrt {{{{k_{eq}}} \over {m'}}} $$
= $${1 \over {2\pi }}\sqrt {{{2k} \over 8}} $$
= $${1 \over {2\pi }}\sqrt {{{2 	imes 4{\pi ^2}} \over 8}} $$
= $${1 \over {2\pi }} 	imes \pi $$
= $${1 \over 2}$$ Hz

Answer

$${1 \over 4}Hz$$

Answer

$${1 \over {2\sqrt 2 }}Hz$$

Answer

$${1 \over 2}Hz$$

Answer

$$2$$ $$Hz$$

Oscillations

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 :

Choose the correct answer: