The oscillating circuit has its own frequency v = 60 kHz. How many kilohertz will the natural frequency of the circuit

The oscillating circuit has its own frequency v = 60 kHz. How many kilohertz will the natural frequency of the circuit change if the inductance is increased by n = 2.25.

The natural frequency of the oscillating circuit before increasing the inductance: ν1 = 1 / T1 = 1 / (2 * Π * √ (L1 * C)) = 60 kHz.
The natural frequency of the oscillating circuit after increasing the inductance by a factor of 2.25 (L2 = 2.25L1): ν2 = 1 / T2 = 1 / (2 * Π * √ (L2 * C)) = 1 / (2 * Π * √ ( 2.25L1 * C)) =) 1 / √2.25) * ν1 = 0.667 * ν1 = (1 / 1.5) * 30 * 10 ^ 3 = 20 * 10 ^ 3 Hz = 20 kHz.
How much will the frequency change: Δν = ν1 – ν2 = 30 kHz – 20 kHz = 10 kHz.
Answer: The natural frequency of the loop will decrease by 10 kHz.