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?

Question

Answer

F=1/4 hz

Answer

F =1 by 4 =0.25

Answer

0.25Hz

Answer

F1=1/2π×√k/m,f2=1/2π√2k/8m=1/2π√k/4mF2=F1/2

Answer

Analysis:

The frequency of vibration of a block attached to a spring is given by the formula:

$$ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $$

where
- $ f $ is the frequency of vibration
- $ k $ is the spring constant
- $ m $ is the mass of the block

Given:

For the 1 kg block:
- $ m = 1 \, \text{kg} $
- $ f = 1 \, \text{Hz} $

Calculations:
The spring constant for the 1 kg block can be found using the given frequency:

$$ 1 = \frac{1}{2\pi} \sqrt{\frac{k}{1}} $$
$$ k = 4\pi^2 \, \text{N/m} $$

Frequency for the 8 kg block:
When two identical springs are attached in parallel, the effective spring constant is the sum of the individual spring constants:

$$ k_{\text{total}} = 2k = 8\pi^2 \, \text{N/m} $$

For the 8 kg block:
- $ m = 8 \, \text{kg} $

Using the formula for frequency of vibration:

$$ f' = \frac{1}{2\pi} \sqrt{\frac{k_{\text{total}}}{m}} $$
$$ f' = \frac{1}{2\pi} \sqrt{\frac{8\pi^2}{8}} $$
$$ f' = \frac{1}{2\pi} \sqrt{\pi^2} $$
$$ f' = \frac{1}{2\pi} \pi $$
$$ f' = 0.5 \, \text{Hz} $$

Therefore, the frequency of vibration of the 8 kg block will be 0.5 Hz when attached to two springs in parallel.

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