All questions of Simple Harmonic Motion (SHM) and Oscillations for JEE Exam

By the time bigger pendulum completes one vibration, the smaller pendulum would have completed 5/4 vibrations. That is smaller pendulum will be ahead by 1/4 vibration in phase. 1/4 vibration means λ/4 path or π/2 radians.

Frequency used to be measured in cycles per second, but now we use the unit of frequency - the Hertz (abbreviated Hz). One Hertz (1Hz) is equal to one vibration per second. So the weight above is bouncing with a frequency of about 1Hz. The sound wave corresponding to Middle C on a piano is around 256Hz.

+ 2) meters.

We can find the velocity and acceleration functions by taking the first and second derivatives of x with respect to time:

velocity v(t) = dx/dt = -3(0.45)sin(0.45t + 2)

acceleration a(t) = d^2x/dt^2 = -3(0.45)^2cos(0.45t + 2)

To find the maximum velocity, we set v(t) equal to zero and solve for t:

0 = -3(0.45)sin(0.45t + 2)

sin(0.45t + 2) = 0

0.45t + 2 = nπ, where n is an integer

t = (nπ - 2)/0.45

To find the maximum acceleration, we set a(t) equal to zero and solve for t:

0 = -3(0.45)^2cos(0.45t + 2)

cos(0.45t + 2) = 0

0.45t + 2 = (n + 0.5)π, where n is an integer

t = [(n + 0.5)π - 2]/0.45

Note that there are infinitely many solutions for both t_max_v and t_max_a, as there are infinitely many values of n.

To find the values of maximum velocity and maximum acceleration, we can substitute the values of t_max_v and t_max_a into the corresponding velocity and acceleration functions:

v_max = -3(0.45)sin(0.45t_max_v + 2)

a_max = -3(0.45)^2cos(0.45t_max_a + 2)

However, since there are infinitely many solutions for t_max_v and t_max_a, we cannot find a unique value for v_max and a_max without additional information.

In an ideal environment where there is no resistance to oscillatory motion, that is, damping is zero, when we oscillate a system at its resonant frequency, since there is no opposition to oscillation, a very large value of amplitude will be recorded. Forced oscillation is when you apply an external oscillating force.

Understanding the Function
The function given is sin(ωt) + cos(ωt). To determine the period of this function, we need to analyze its components.
Components of the Function
- The sine function, sin(ωt), has a standard period of 2π.
- The cosine function, cos(ωt), also has a period of 2π.
Since both functions share the same frequency, the overall period of their sum will also be 2π.
Calculating the Period
1. Period of Sine Function:
- The period is given by T = 2π/ω.
2. Period of Cosine Function:
- Similarly, the period is also T = 2π/ω.
Since both functions oscillate with the same frequency, the resultant function sin(ωt) + cos(ωt) will also have a period of 2π/ω.
Correct Option
Thus, the correct answer for the time period of the function sin(ωt) + cos(ωt) is:
- Option B: 2π/ω
This means that every 2π/ω seconds, the function will complete one full cycle.
Conclusion
In summary, the periodic behavior of both sine and cosine functions, when combined, maintains the same period of 2π/ω. Therefore, recognizing these fundamental properties allows us to confidently determine the period of the combined function.

**Explanation:**

When a particle is executing Simple Harmonic Motion (SHM), its displacement from the mean position follows a sinusoidal function. The maximum displacement of the particle is called the amplitude (A) of the motion. The particle oscillates back and forth between two extreme positions, which are equal in magnitude but opposite in direction.

The time taken for one complete oscillation of the particle is called the time period (T) of the motion. It is the time taken for the particle to go from one extreme position to the other and back again.

The maximum velocity (vmax) of the particle occurs when the particle passes through the mean position (equilibrium position). At this point, the velocity is at a maximum and is in the same direction as the acceleration. The maximum acceleration (amax) of the particle occurs when the particle is at its extreme positions. At these points, the acceleration is at a maximum and is in the opposite direction to the displacement.

If the maximum velocity and acceleration of the particle are equal in magnitude, it means that the amplitude of the motion is equal to the acceleration. Mathematically, this can be represented as:

amax = vmax

Using the equations of SHM, we can relate the amplitude, time period, maximum velocity, and maximum acceleration as follows:

amax = ω^2 A

vmax = ω A

where ω is the angular frequency of the motion.

Since amax = vmax, we can equate the above two equations:

ω^2 A = ω A

Simplifying the equation:

ω^2 = ω

ω = 1

The angular frequency ω is related to the time period T by ω = 2π/T. Substituting this value into the equation:

(2π/T)^2 = 2π/T

Simplifying the equation:

4π^2/T^2 = 2π/T

2π/T = 4π^2/T^2

T = 2

Therefore, the time period of the motion is T = 2 seconds, which is equivalent to 6.28 seconds (approximately). Hence, the correct answer is option A: 6.28 s.

Harmonic motion is the natural motion of a body(we consider no air friction) under no force where as damped oscillation are under force hence the iscilation are different