A mass of m = 0.250 kg is attached to the end of a spring of force constant k = 40 N m^-1 and hangs in equilibrium. The mass is then pulled down a further 6.0 cm from its equilibrium position and released from rest at time t=0, so that it oscillates with simple harmonic motion o…

Answer: ((a)) x=x0 t; the graph is a smooth cosine curve starting at +x0 at t=0, falling through zero at t=T/4, reaching -x0 at t=T/2, back to zero at t=3T/4, and returning to +x0 at t=T, repeating periodically with amplitude x0 and period T ((b)) =√(k/m)=√(40/0.250)=√(160)=12.6 rad s^-1 (3 s.f.) ((c)) vmax=0.759 m s^-1 (3 s.f.) ((d)) amax=9.60 m s^-2. Part (a): Displacement equation and shape of the graph The mass is pulled down to its maximum displacement (x=x0) and released from rest. Since displacement is a maximum and velocity is zero at t=0, the appropriate solution of the SHM equa…

A Level Physics 9702: SHM of a Mass on a Spring - Period, Speed and Acceleration

A mass of $m = 0.250\text{ kg}$ is attached to the end of a spring of force constant $k = 40\text{ N m}^{-1}$ and hangs in equilibrium. The mass is then pulled down a further $6.0\text{ cm}$ from its equilibrium position and released from rest at time $t=0$ , so that it oscillates with simple harmonic motion of amplitude $x_0 = 0.060\text{ m}$ .

(a) State the equation for the displacement $x$ of the mass at time $t$ after release, and describe the shape of the graph of $x$ against $t$ that this equation represents over one complete oscillation. [2]

(b) Show that the angular frequency $\omega$ of the oscillation is $12.6\text{ rad s}^{-1}$ . [2]

(d) Calculate the maximum magnitude of the acceleration of the mass during the oscillation. [2]

Part (a): Displacement equation and shape of the graph

The mass is pulled down to its maximum displacement ( $x=x_0$ ) and released from rest. Since displacement is a maximum and velocity is zero at $t=0$ , the appropriate solution of the SHM equation is a cosine, not a sine: $x=x_0\cos\omega t$

(By contrast, $x=x_0\sin\omega t$ describes a particle that starts at $x=0$ moving with maximum speed, not the case here.)

Shape of the graph: the graph of $x$ against $t$ is a smooth, continuous cosine wave. It starts at its maximum value $+x_0$ when $t=0$ , decreases smoothly through zero displacement at $t=T/4$ , reaches its minimum value $-x_0$ at $t=T/2$ , rises back through zero at $t=3T/4$ , and returns to $+x_0$ at $t=T$ , repeating with this same shape indefinitely, with amplitude $x_0$ and period $T$ (this idealised case ignores any damping).

Part (b): Show that ω = 12.6 rad s⁻¹

For a mass-spring system, $T=2\pi\sqrt{\dfrac{m}{k}}$ . Since $\omega=\dfrac{2\pi}{T}$ , substituting gives: $\omega=\frac{2\pi}{2\pi\sqrt{m/k}}=\frac{1}{\sqrt{m/k}}=\sqrt{\frac{k}{m}}$

Substituting the values: $\omega=\sqrt{\frac{40}{0.250}}=\sqrt{160}=12.649\text{ rad s}^{-1}$

Check by recomputing separately: $12.649^2=160.0$ , and $k/m=40/0.250=160$ , the two agree, confirming $\omega=\sqrt{160}$ .

So $\omega=12.6\text{ rad s}^{-1}$ (3 s.f.), as required to show.

Part (c): Maximum speed

$v_{max}=\omega x_0=12.649\times0.060=0.759\text{ m s}^{-1}\ (3\text{ s.f.})$

Recompute as a check: $0.060\times12.65=0.759\text{ m s}^{-1}$ . Consistent.

Part (d): Maximum acceleration

$a_{max}=\omega^2x_0=(12.649)^2\times0.060=160\times0.060=9.60\text{ m s}^{-2}$

Recompute independently using $a_{max}=\dfrac{k}{m}x_0$ directly (since $\omega^2=k/m$ ): $\dfrac{40}{0.250}\times0.060=160\times0.060=9.60\text{ m s}^{-2}$ . Both routes agree exactly.

Final answers

(a) $x=x_0\cos\omega t$ ; a cosine-shaped graph starting at $+x_0$ , passing through $0$ at $T/4$ , reaching $-x_0$ at $T/2$ , and repeating with period $T$
(b) $\omega=\boxed{12.6}\text{ rad s}^{-1}$
(c) $v_{max}=\boxed{0.759}\text{ m s}^{-1}$
(d) $a_{max}=\boxed{9.60}\text{ m s}^{-2}$