Two identical springs of constant are connected in series and parallel as shown in figure. A mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

Question

Two identical springs of constant are connected in series and parallel as shown in figure- A mass m is suspended from them- The ratio of their frequencies of vertical oscillations will be1-22-11-14-1

Toppr Admin · Accepted Answer

We know that- n-12-x3C0-x221A-kmIn series- k-k1k2k1-k2-k22In parallel- k-k1-k2-2k-x2234-n1-12-x3C0-x221A-k2mand n2-12-x3C0-x221A-2kmhence- n1-n2-1-2

Two identical springs of constant are connected in series and parallel as shown in figure. A mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

$1 : 2$
$2 : 1$
$1 : 1$
$4 : 1$

We know that, $n = \frac{1}{2 π} \sqrt{\frac{k}{m}}$
In series, $k = \frac{k_{1} k_{2}}{k_{1} + k_{2}} = \frac{k^{2}}{2}$
In parallel, $k = k_{1} + k_{2} = 2 k$
$∴, n_{1} = \frac{1}{2 π} \sqrt{\frac{k}{2 m}}$
and $n_{2} = \frac{1}{2 π} \sqrt{\frac{2 k}{m}}$
hence, $n_{1} : n_{2} = 1 : 2$

Two identical springs of constant are connected in series and parallel as shown in figure. A mass m is suspended from them. The ratio of their frequencies of vertical oscillations will be1:22:11:14:1

We know that, n=12π√kmIn series, k=k1k2k1+k2=k22In parallel, k=k1+k2=2k∴,n1=12π√k2mand n2=12π√2kmhence, n1:n2=1:2

Two identical springs of constant are connected in series and parallel as shown in figure. A mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

$1 : 2$
$2 : 1$
$1 : 1$
$4 : 1$

We know that, $n = \frac{1}{2 π} \sqrt{\frac{k}{m}}$
In series, $k = \frac{k_{1} k_{2}}{k_{1} + k_{2}} = \frac{k^{2}}{2}$
In parallel, $k = k_{1} + k_{2} = 2 k$
$∴, n_{1} = \frac{1}{2 π} \sqrt{\frac{k}{2 m}}$
and $n_{2} = \frac{1}{2 π} \sqrt{\frac{2 k}{m}}$
hence, $n_{1} : n_{2} = 1 : 2$