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Chapter 1: Q26E (page 461)

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.250 m and the period is 3.20 s. What are the speed and acceleration of the block when x = 0.160 m?

Short Answer

Expert verified

The speed and acceleration of the block isvx=0.377m/s and ax=0.62m/s2.

Step by step solution

01

Formula required for velocity and angular frequency

Velocity at any point x is given as,

Vx=卤KmA2-x2 鈥︹�︹��.. (1)

Where K is the spring constant, A is the amplitude of the SHM, and x denotes position of object.

We know that angular frequency is given as,

蝇=Km=2螤T 鈥︹�︹��.. (2)

02

Calculate speed and acceleration      

From equation (1) and (2)

Vx=蝇A2鈭�x2=2螤TA2鈭�x2 鈥︹�︹��.. (3)

The amplitude of SHM is A = 0.250 m and time period is T = 3.20 sec.

The position of block is x=0.160 m,

Hence, from equation (3)

Vx=2螤3.2s(0.25m)2鈭�(0.16m)2=0.377m/s

We know acceleration is given as,

ax=鈭�蝇2x=2螤3.2s2脳(0.16m)=0.62m/s2

At x=0.16 m speed is 0.377 m/s and acceleration is 0.62m/s2

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