Q.A body oscillates with SHM according to the equation (in SI units), $x = 5\cos\left[2\pi t + \pi/4\right]$. At $t = 1.5$ s, calculate the

The key idea is that for SHM given by $x = A\cos(\omega t + \phi)$, the velocity is $v = -\omega A\sin(\omega t + \phi)$ and acceleration is $a = -\omega^2 x$.

Step 1 — Identify parameters.

From $x = 5\cos(2\pi t + \pi/4)$, we have amplitude $A = 5$ m, angular frequency $\omega = 2\pi$ rad/s, and phase constant $\phi = \pi/4$ rad.

Step 2 — Displacement at $t = 1.5$ s.

$x = 5\cos(2\pi \times 1.5 + \pi/4) = 5\cos(3\pi + \pi/4) = 5\cos(13\pi/4)$.

Since $\cos(13\pi/4) = \cos(5\pi/4) = -\frac{\sqrt{2}}{2}$, we get

$x = 5 \times (-\sqrt{2}/2) = -\frac{5\sqrt{2}}{2}$ m.

Step 3 — Speed and acceleration.

$v = -\omega A\sin(\omega t + \phi) = -2\pi \times 5 \times \sin(13\pi/4)$. …

Why This Approach Works

Simple Harmonic Motion is described by sinusoidal functions because the restoring force is proportional to displacement. The equation $x = A\cos(\omega t + \phi)$ is the most general solution — here $A = 5$ m is the amplitude, $\omega = 2\pi$ rad/s is the angular frequency, and $\phi = \pi/4$ rad is the phase constant.

The key insight: once you know the displacement as a function of time, velocity and acceleration come from differentiation, not from memorising separate formulas. This is because velocity is the rate of change of position, and acceleration is the rate of change of velocity. For SHM, differentiating a cosine gives a negative sine for velocity, and differentiating again gives a negative cosine for acceleration — which is proportional to displacement itself.

Let's work through each part systematically.

1. Find the displacement at $t = 1.5$ s

The displacement equation is:

Substitute $t = 1.5$:

Now $3\pi = \pi + 2\pi$, so:

Since cosine has period $2\pi$, we can drop the $2\pi$ term:

Using the identity $\cos(\pi + \theta) = -\cos\theta$:

Therefore:

2. Find the velocity at $t = 1.5$ s

Velocity is the first derivative of displacement with respect to time:

Using the chain rule: derivative of $\cos$ is $-\sin$, multiplied by the derivative of the inside ($2\pi$):

Now substitute $t = 1.5$:

Again, reduce the angle: $\sin(\pi + \theta) = -\sin\theta$, so:

Thus:

Numerically, $\pi \approx 3.1416$, so:

The speed is the magnitude of velocity, so speed $= 22.21$ m/s. …

Step 1: Identify the SHM parameters from $x=5\cos(2\pi t+\pi/4)$: amplitude $A=5\ \text{m}$, angular frequency $\omega=2\pi\ \text{rad/s}$, phase $\phi=\pi/4$.

Step 2: At $t=1.5\ \text{s}$ the phase is $\omega t+\phi = 3\pi+\pi/4=13\pi/4$, which is coterminal (mod $2\pi$) with $5\pi/4$.

Step 3: Displacement: $x=5\cos(5\pi/4)=5\times(-\sqrt2/2)=-\dfrac{5\sqrt2}{2}\approx-3.54\ \text{m}$. …