JEE Advanced 2013 · Paper 2 · Physics Q6

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PhysicsMultiple correct+3 / -1Question 6 of 60

A particle of mass $m$ is attached to one end of a mass-less spring of force constant $k$ , lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$ . When the speed of the particle is $0.5u_0$ , it collides elastically with a rigid wall. After this collision,

the speed of the particle when it returns to its equilibrium position is

u_0

the time at which the particle passes through the equilibrium position for the first time is

t=\pi\sqrt{\dfrac mk}

the time at which the maximum compression of the spring occurs is

t=\dfrac{4\pi}3\sqrt{\dfrac mk}

the time at which the particle passes through the equilibrium position for the second time is

t=\dfrac{5\pi}3\sqrt{\dfrac mk}

Answer and worked solution

Official answer: A, D

Motion. $x=\dfrac{u_0}\omega\sin\omega t$ with $\omega=\sqrt{k/m}$ . The speed is $0.5u_0$ when $\cos\omega t=\frac12$ , i.e. $\omega t_1=\frac\pi3$ .

After the elastic bounce the velocity reverses, so the particle retraces its motion:

back at equilibrium after another $\frac\pi{3\omega}$ , i.e. $t=\frac{2\pi}{3\omega}$ , with speed $u_0$ ✓ (A); (B) is wrong;
maximum compression a quarter period later: $t=\frac{2\pi}{3\omega}+\frac\pi{2\omega}=\frac{7\pi}{6\omega}$ , so (C) is wrong;
back at equilibrium half a period after the first return: $t=\frac{2\pi}{3\omega}+\frac\pi\omega=\frac{5\pi}{3\omega}$ ✓ (D).

(A), (D)

Topics:Simple harmonic motion Spring-mass systems

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JEE Advanced 2013 · Paper 2, Physics question 6