Question

Two block of masses m1 = 2kg and m2 = 8 kg are connected by
a spring of force constant K = 1000 kg/m. If the blocks are
caused to oscillate, the frequency 'f' of motion in Hertz is
A. 25√2π
B. 25/2π
C. 25/π√2π
D. 2π/25​

Answer 1

Answer:

• The frequency (f) of the motion is 25 / 2 π Hz

Given:

1. Mass of first body (M₁) = 2 Kg
2. Mass of second body (M₂) = 8 Kg
3. Force Constant (K) = 1000 N/m

Explanation:

$\rule{300}{1.5}$

As two blocks are attached to a spring, therefore it is a case of Reduced mass.

Finding the reduced mass (μ)

From, the formula we know,

$\bigstar\;\underline{\boxed{\sf \mu=\dfrac{M_1\;M_2}{M_1+M_2}}}$

Here,

• M₁ Denotes First body mass.
• M₂ Denotes Second body mass.

Substituting the values,

$\displaystyle\longrightarrow\sf \mu =\dfrac{2\times 8}{2+8}\\\\\\\longrightarrow\sf \mu =\dfrac{16}{10}\\\\\\\longrightarrow\sf \mu=1.6\\\\\\\longrightarrow \underline{\boxed{\sf \mu=1.6 }}$

∴ We got the reduced mass.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, from the formula we know,

$\bigstar\;\underline{\boxed{\sf f=\dfrac{1}{T}=\dfrac{1}{2\;\pi}\sqrt{\dfrac{K}{\mu}}}}$

Here,

• K Denotes Force constant.
• μ denotes Reduced mass.

Substituting the values,

$\displaystyle\longrightarrow\sf f=\dfrac{1}{2\;\pi}\sqrt{\dfrac{1000}{1.6}}\\\\\\\longrightarrow\sf f=\dfrac{1}{2\;\pi}\sqrt{\dfrac{10000}{16}}\\\\\\\longrightarrow\sf f=\dfrac{1}{2\;\pi}\times \dfrac{100}{4}\\\\\\\longrightarrow\sf f=\dfrac{1}{2\;\pi}\times 25\\\\\\\longrightarrow\sf f=\dfrac{25}{2\;\pi}\\\\\\\longrightarrow\sf \large{\underline{\boxed{\red{\sf f=\dfrac{25}{2\;\pi}\;Hz}}}}$

∴ The frequency (f) of the motion is 25 / 2 π Hz.

Hence Option-B is correct !

$\rule{300}{1.5}$

Answer 2

$\huge\underline\mathtt\red{Answer:-}$

•The frequency (f) of the motion is 25 / 2 π Hz