Answer
The new value of $\omega$ is $5.0~rad/s$
Work Step by Step
Let the equivalent spring constant of the two springs be $k$. Let $M$ be the mass of the cart. We can find an expression for the original value of $\omega$:
$\omega = \sqrt{\frac{k}{M}}$
Let $M' = 4.0~M$. We can find an expression for the new value of the angular frequency $\omega'$:
$\omega' = \sqrt{\frac{k}{M'}}$
$\omega' = \sqrt{\frac{k}{4.0~M}}$
$\omega' = \frac{1}{2.0}\times \sqrt{\frac{k}{M}}$
$\omega' = \frac{1}{2.0}\times \omega$
$\omega' = \frac{1}{2.0}\times 10.0~rad/s$
$\omega' = 5.0~rad/s$
The new value of $\omega$ is $5.0~rad/s$.
