Two springs are connected to a block of mass M placed on a frictionless surface as shown below. If both the springs have a spring constant k, then the frequency of oscillation of the block is

Question

Two springs are connected to a block of mass M placed on a frictionless surface as shown below. If both the springs have a spring constant k, then the frequency of oscillation of the block is (A) (1/2 π) √(k/M) (B) (1/2 π) √(k/2 M) (C) (1/2 π) √(2 k/M) (D) (1/2 π) √(M/k)

Tardigrade Admin · Accepted Answer

The frequency of oscillation of a vibrating spring-block, system is given by

v = \frac{1}{2 π} \frac{k _{eff}}{M} \dots (i)

Since, the springs are attached in series, each having spring constant

k

, so we have

\frac{1}{k _{e ff}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}

\Rightarrow \frac{1}{k _{e ff}} = \frac{k}{2}

From Eq. (i), we get

v = \frac{1}{2 π} \frac{k _{eff}}{M} = \frac{1}{2 π} \frac{k}{2 M}

Q. Two springs are connected to a block of mass $M$ placed on a frictionless surface as shown below. If both the springs have a spring constant $k$ , then the frequency of oscillation of the block is

$\frac{1}{2 π} \frac{k}{M}$

$\frac{1}{2 π} \frac{k}{2 M}$

$\frac{1}{2 π} \frac{2 k}{M}$

$\frac{1}{2 π} \frac{M}{k}$

Q. Two springs are connected to a block of mass M placed on a frictionless surface as shown below. If both the springs have a spring constant k, then the frequency of oscillation of the block is

2π1​Mk​​

2π1​2Mk​​

2π1​M2k​​

2π1​kM​​

Q. Two springs are connected to a block of mass $M$ placed on a frictionless surface as shown below. If both the springs have a spring constant $k$ , then the frequency of oscillation of the block is

$\frac{1}{2 π} \frac{k}{M}$

$\frac{1}{2 π} \frac{k}{2 M}$

$\frac{1}{2 π} \frac{2 k}{M}$

$\frac{1}{2 π} \frac{M}{k}$