{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

442_Physics ProblemsTechnical Physics

# 442_Physics ProblemsTechnical Physics - 444*P15.8...

This preview shows page 1. Sign up to view the full content.

444 Oscillatory Motion *P15.8 The mass of the cube is mV == × = × ρ 27 10 0015 911 10 3 3 3 .. . kg m m kg 3 ej af The spring constant of the strip of steel is k F x f k m = = × = 14 3 52 0 2 1 2 1 2 52 12 0 3 . . . . N 0.027 5 m Nm kg s k g Hz 2 ω ππ π P15.9 f k m 2 1 2 or T f m k 1 2 Solving for k , k m T = 4 47 0 0 260 40 9 2 2 2 2 . . . kg s ±Nm bg a f . *P15.10 xA t = cos A = 005 . m vA t =− ωω sin aA t 2 cos If f 3600 60 rev min Hz, then ωπ = 120 1 s v max 005120 188 ms a max 711 2 ±ms ±kms 22 P15.11 (a) = k m 800 0500 400 1 . . . kg s so position is given by xt = 10 0 4 00 .s i n . cm. From this we find that vt = 40 0 4 00 .c o s . cm s v max . = 40 0 cm s at 160 4 00 sin . cm s 2 a max = 160 cm s 2 . (b) t x = F H G I K J F H G I K J 1 100 1 . sin . and when x = 600 c m , t = 0161
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online