Answer to Question #275009 in Physics for Imu

Question

The equation of a particle executing in SHM is given by,y=24Sin(omegat+phi) , if

the time period is 30 second and the particle has a displacement of 12cm at t=0

, Calculate (i) epoch, (ii) the phase angle at t=10 secods and iii) the

phase difference between two positions of the particle “S” seconds apart

(where S is the last 2 digits of your student ID except invalid 0).

Answer 1

The equation:

$$y=24\sin(\omega t+\phi).$$

We know that

$$\omega=\frac{2\pi}T.\\\space\\ y=24\sin\bigg(\frac{2\pi}Tt+\phi\bigg).$$

(i) Epoch:

$$12=24\sin\bigg(\frac{2\pi}T·0+\phi\bigg),\\\space\\ \phi=\frac \pi6.$$

(ii) The phase angle at t=10 s:

$$\frac{2\pi}T·10+\phi=\frac {5\pi}6$$

because 10 s are less than the period.

iii) The phase difference between two positions of the particle “22” seconds apart:

$$\Delta\chi=\frac{2\pi}T·22=4.61\text{ rad}.$$