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A particle pe...

Question

A particle performing $S H M$ with frequency $10 H z$ and amplitude $5 c m$ is initially in left extreme position. The equation of its displacement will be ( $x$ is in metre)

x = 0.05 s i n (20 π t + \frac{3 π}{2})

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x = 0.05 s i n (20 π t + \frac{5 π}{2})

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x = 0.05 s i n (10 π t)

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x = 0.05 s i n (20 π t + π)

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Solution

The correct option is D $x = 0.05 s i n (20 π t + π)$
In general, $x = A s i n (ω t + ϕ)$
Amplitude $A = 5 c m$
frequency $f = 10$
$\Rightarrow ω = 2 π f$
$ω = 20 π$
$\Rightarrow x = 5 c m s i n (20 π t + ϕ)$
Since $x = - 5 c m$ at $t = 0$
$- 5 = 5 s i n ϕ$
$s i n ϕ$ = -1
$\Rightarrow ϕ = 3 π / 2$
$∴ x = 5 c m s i n (20 π t + 3 π / 2)$
As $x$ is in meters
$∴ x = 0.05 s i n (20 π t + 3 π / 2)$

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A particle performing SHM with frequency 10 Hz and amplitude 5 cm is initially in left extreme position. The equation of its displacement will be (x is in metre)

The correct option is D x=0.05 sin (20π t+π)In general, x=A sin(ωt+ϕ) Amplitude A=5 cm frequency f=10 ⇒ω=2π f ω=20π ⇒x=5 cm sin(20π t+ϕ) Since x=−5 cm at t=0 −5=5 sin ϕ sin ϕ= -1 ⇒ϕ=3π/2 ∴x=5 cm sin(20 πt+3π/2) As x is in meters ∴x=0.05 sin (20π t+3π/2)

A particle performing $S H M$ with frequency $10 H z$ and amplitude $5 c m$ is initially in left extreme position. The equation of its displacement will be ( $x$ is in metre)