The displacement $x$ (in metres) of a particle performing simple harmonic motion is related to time $t$ (in seconds as} $x = 0.05 cos (4 π t + \frac{π}{4})$ . The frequency of the motion will be

0.5

H z

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1.0

H z

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1.5

H z

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2.0

H z

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Solution

The correct option is D $2.0$ $H z$
Comparing the given equation with the standard equation of SHM i.e. $y = a s i n (ω t + ϕ)$

$y = 0.05 s i n (\frac{π}{2} - (4 π t + \frac{π}{4}))$

$y = 0.05 s i n (\frac{π}{4} - 4 π t)$

So, the angular frequency of the oscillation will be:
$ω = 4 π$

So, frequency of oscillation will be:
$f = \frac{ω}{2 π}$

$f = \frac{4 π}{2 π} = 2 H z$

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Similar questions

The displacement x (in metre) of a particle in, simple harmonic motion is related to time t (in seconds) as

x = 0.01 cos $(π t + \frac{π}{4})$

The frequency of the motion will be

0.01 c o s (π t + \frac{π}{4})

x = - 0.3 s i n (t + \frac{π}{4})

2 cm

1 cm

10 H z

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