Question

problem 1
an intel core processor has a frequency of 4.8 ghz (number of possible in - structions per second). find the period in seconds and the angular frequency in rad/s.

Explanation:

Step1: Convert GHz to Hz

Given frequency \( f = 4.8\ GHz=4.8\times10^{9}\ Hz\)

Step2: Calculate the period \(T\)

The formula for the period is \(T=\frac{1}{f}\).
Substituting \(f = 4.8\times 10^{9}\ Hz\), we get \(T=\frac{1}{4.8\times 10^{9}}=\frac{1}{4.8}\times10^{- 9}\approx2.08\times10^{-10}\ s\)

Step3: Calculate the angular frequency \(\omega\)

The formula for angular frequency is \(\omega = 2\pi f\).
Substituting \(f = 4.8\times 10^{9}\ Hz\), we get \(\omega=2\pi\times4.8\times 10^{9}=9.6\pi\times10^{9}\approx3.016\times 10^{10}\ rad/s\)

Answer:

The period \(T\approx2.08\times 10^{-10}\ s\) and the angular frequency \(\omega\approx3.016\times 10^{10}\ rad/s\)