Assuming the particle to be stationary at the moment $t = 0$ , find its velocity as a function of time.

| v (t) | = (\frac{2 \to F}{m ω}) ∣ ∣ ∣ sin (\frac{ω t}{2}) ∣ ∣ ∣

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| v (t) | = (\frac{\to F}{m ω}) ∣ ∣ ∣ sin (\frac{ω t}{2}) ∣ ∣ ∣

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| v (t) | = (\frac{2 \to F}{2 m ω}) ∣ ∣ ∣ sin (\frac{ω t}{2}) ∣ ∣ ∣

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| v (t) | = 2 (\frac{2 \to F}{m ω}) ∣ ∣ ∣ sin (\frac{ω t}{2}) ∣ ∣ ∣

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Solution

The correct option is A $| v (t) | = (\frac{2 \to F}{m ω}) ∣ ∣ ∣ sin (\frac{ω t}{2}) ∣ ∣ ∣$
Given $¯ F = F (c o s (ω t)^i + s i n (ω t)^j)$
So, $v (t) = \int_{0}^{t} \frac{F}{m} (c o s (ω t) ˆ i + s i n (ω t) ˆ j) d t$
$v (t) = \frac{2 F}{m ω} (s i n (ω t) ˆ i + (1 - c o s (ω t)) ˆ j)$
$| v (t) | = \frac{2 F}{m ω} | s i n (\frac{ω t}{2}) |$

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Assuming the particle to be stationary at the moment t=0, find its velocity as a function of time.

The correct option is A |v(t)|=(2→Fmω)∣∣∣sin(ωt2)∣∣∣Given ¯F=F(cos(ωt)^i+sin(ωt)^j)So, v(t)=∫t0Fm(cos(ωt)ˆi+sin(ωt)ˆj)dtv(t)=2Fmω(sin(ωt)ˆi+(1−cos(ωt))ˆj)|v(t)|=2Fmω|sin(ωt2)|

Assuming the particle to be stationary at the moment $t = 0$ , find its velocity as a function of time.