Question

The position of a 55 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.40s. express your answer in meters per second to two significant figures.

Answer 1

The velocity of the oscillating particle at $t=0.4\,{\text{s}}$  is $\boxed{1.5\,{{{\text{cm}}}\mathord{\left/{\vphantom{{{\text{cm}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$  or $\boxed{1.5\times{{10}^{-2}}\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ .

Further Explanation:

The position of the oscillating mass is given by:

$x\left(t\right)=\left({2.0\,{\text{cm}}}\right)\cos\left({10t}\right)$

Here, $x\left(t\right)$  is the position of the particle at time $t$  during the oscillation.

The velocity of the oscillating particle is defined as the rate of change of the position of the body. Thus, it can be expressed as the first derivative of the position of the body while it is oscillating.

The velocity of the particle can be expressed as:

$\boxed{v=\frac{{dx\left(t\right)}}{{dt}}}$

Substitute the equation of the position in above expression.

$\begin{aligned}v&=\frac{d}{{dt}}\left({\left({2.0\,{\text{cm}}}\right)\cos\left({10t}\right)}\right)\\&=-\left({2.0\,{\text{cm}}}\right)\sin\left({10t}\right)\\\end{aligned}$

Now, we are to obtain the velocity of the oscillating particle at time $t=0.4\,{\text{s}}$ . So, substitute $0.4$  for  $t$ in above equation of velocity.

$\begin{aligned}v&=-\left({2.0\,{\text{cm}}}\right)\sin\left({10\times0.4\,{\text{rad}}}\right)\\&=-2.0\times\left({-0.75}\right)\\&=1.5\,{{{\text{cm}}}\mathord{\left/{\vphantom{{{\text{cm}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The velocity of the oscillating particle in ${{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ while it oscillates is given as:

$\begin{aligned}v&=1.5\,{{{\text{cm}}}\mathord{\left/{\vphantom{{{\text{cm}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\left({\frac{{1\,{\text{m}}}}{{100\,{\text{cm}}}}}\right)\\&=1.5\times{10^{-2}}\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

Thus, the velocity of the oscillating particle at $t=0.4\,{\text{s}}$  is $\boxed{1.5\,{{{\text{cm}}}\mathord{\left/{\vphantom{{{\text{cm}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ or $\boxed{1.5\times{{10}^{-2}}\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ .

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Answer Details:

Grade: College

Subject: Physics

Chapter: Oscillation

Keywords:

Position, 55g particle9t, oscillating mass, velocity at, t=0.40 s, position of particle, rate of change of position, x(t)=(2.0 cm)cos(10t).

Answer 2

The position of the mass is given by (in cm):
$x(t)=2 \cos (10 t)$
The velocity is the derivative of the position:
$v(t) = \frac{dx(t)}{dt} =-10\cdot 2 \sin (10t)=-20 \sin (10t)$
Substituting t=0.40 s, we can find the velocity at this time:
$v(0.40 s)= -20 \sin (10 \cdot 0.4)=15 cm/s=15 \cdot 10^{-2}m/s$