Physics · Oscillations · NEET
Use v = w * root(A^2 - x^2). At the mean position x = 0, so v = w * root(A^2) = wA, the maximum. At the extreme position x = A, so v = w * root(A^2 - A^2) = 0. Physically the particle slows down as it climbs toward an extreme, stops for an instant to turn around, then speeds up again toward the middle.
Start with x = A sin(wt + phi) and v = wA cos(wt + phi). Square both: x^2 = A^2 sin^2(wt+phi) and v^2 = w^2 A^2 cos^2(wt+phi). Since sin^2 + cos^2 = 1, we get cos^2(wt+phi) = 1 - x^2/A^2. Substitute: v^2 = w^2 A^2 (1 - x^2/A^2) = w^2 (A^2 - x^2). Take the root: v = w * root(A^2 - x^2). This form is useful when time is not given but position is.
Velocity is a vector; over one full vibration the particle returns to its start, so total displacement is zero and average velocity = 0. Speed is the magnitude; the particle covers a path length of 4A in one period, so average speed = 4A / T, which is not zero. NEET 2019 asked exactly this and the answer is zero (average velocity).
The sign of v = wA cos(wt + phi) just tells the direction of motion. Moving away from the mean position toward the +x extreme, v is positive; coming back it is negative. For magnitude questions use |v| = w * root(A^2 - x^2), which is always positive.
If x = A sin(wt), then v = wA cos(wt) = wA sin(wt + pi/2). The extra pi/2 means velocity reaches its peak a quarter cycle before displacement. When displacement is zero (mean position) velocity is at its maximum, and when displacement is maximum (extreme) velocity is zero. That quarter-cycle shift is the phase difference of pi/2.
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity equals the magnitude of its acceleration. Its time period (in seconds) is
The average velocity of a particle executing SHM over one complete vibration is
Two identical point masses P and Q, suspended from two separate massless springs of spring constants k1 and k2 respectively, oscillate vertically. If their maximum speeds are the same, the ratio A_Q/A_P of the amplitude of mass Q to that of mass P is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
In terms of time, v = w * A * cos(wt + phi). In terms of position, v = w * root(A^2 - x^2). Here w is the angular frequency, A the amplitude, and x the displacement from the mean position.
The maximum speed is v_max = wA and it occurs at the mean position where x = 0. As the particle moves out to x = A the speed drops to zero.
Velocity is zero at the extreme positions, where x = A (or x = -A). At these turning points the particle stops for an instant and reverses direction.
Velocity leads displacement by a phase of pi/2 (90 degrees). When displacement is zero the velocity is maximum, and when displacement is maximum the velocity is zero.
Yes. Over one complete vibration the particle returns to its start, so net displacement is zero and average velocity is zero. Average speed, however, is 4A/T and is not zero.
