Question

The displacement of a particle executing SHM is given by X = 3 sin [2πt + π/4] , where 'X' is in meter and 't' is in second. The amplitude and maximum speed of the particle is

• A. 3 m, 6π ms-1
• B. 3 m, 2π ms-1
• C. 3 m, 8π ms-1
• D. 3 m, 4π ms-1

Answer 1

Answer: (A)

3 m, 6π ms-1

Answer 2

To find the amplitude and maximum speed of the particle, we can analyze the given equation for displacement:
X = 3 sin(2πt + π/4)
Comparing this equation with the general form of SHM, X = A sin(ωt + φ), we can identify the amplitude (A) and angular frequency (ω). Amplitude (A) is the coefficient of the sine function, which in this case is 3.
Amplitude (A) = 3 m
Angular frequency (ω) is the coefficient of 't' inside the sine function, which is 2π.
Angular frequency (ω) = 2π rad/s
Now, the maximum speed of a particle executing SHM occurs when the displacement is at its maximum value. In this case, the maximum displacement is equal to the amplitude (A).
Maximum speed (Vmax) can be calculated using the formula:
Vmax = ωA
Substituting the values we found:
Vmax = (2π rad/s)(3 m)
Vmax = 6π m/s
Therefore, the amplitude and maximum speed of the particle are 3 m and 6π m/s, respectively.
So, the correct option is (A) 3 m, 6π ms-1.