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Question

The point of the needle of a sewing machine moves in SHM along the x -axis with a frequency of 2.5 Hz. At t = 0 its position and velocity components are +1.1 cm and -15 cm/s, respectively. Find the acceleration component of the needle at t = 0.

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Here given that the frequency of the needle of a sewing machine is f=2.5Hzf = 2.5\,\mathrm{Hz}.
At time t=0st = 0\,\mathrm{s}, the position of the needle is x0=+1.1cm=+1.1×102mx_0 = + 1.1\,\mathrm{cm} = + 1.1\times10^{-2}\,\mathrm{m}
and velocityof the needle is v0x=15cms1=15×102ms1v_{0\text{x}} = - 15\,\mathrm{cm\,s^{-1}} = - 15\times10^{-2}\,\mathrm{m\,s^{-1}}. Here subscript x\text{x} shows that the motion is along x-axis.
(a)\textbf{(a)} As we know that the acceleration of the object in SHM is given by

ax=ω2x(1)\begin{aligned} a_{\text{x}} = -\omega^2x\tag{1} \end{aligned}

Here ω=2πf\omega = 2\pi f and x=x0x = x_0, so

ax=4π2f2x0=4π2×(2.5Hz)2×1.1×102m=4π2×(2.5s1)2×1.1×102m=2.71ms2(2)\begin{aligned} a_{\text{x}} & = -4 \pi^2 f^2x_0\\ & = -4 \pi^2 \times \left(2.5\,\mathrm{Hz} \right)^2 \times 1.1\times10^{-2}\,\mathrm{m}\\ & = -4 \pi^2 \times \left(2.5\,\mathrm{s^{-1}} \right)^2 \times 1.1\times10^{-2}\,\mathrm{m}\\ & = -2.71\,\mathrm{m\,s^{-2}} \tag{2} \end{aligned}

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