LIVE Course for free

Rated by 1 million+ students
Get app now
JEE MAIN 2023
JEE MAIN 2023 TEST SERIES
NEET 2023 TEST SERIES
NEET 2023
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
25 views
in Physics by (85.6k points)
closed by
A spring of force constant `1200Nm^(-1)` is mounted on a horizontal table as shown in figure. A mass of 3.0kg is attached to the free end of the spring, pulled side ways to a distance of 2.0cm and released. Determing.
(a) the frequency of oscillation of the mass.
(b) the maximum acceleration of the mass.
(c) the maximum speed of the mass.
image

1 Answer

0 votes
by (86.1k points)
selected by
 
Best answer
Here, `k=1200Nm^(-1),m=3.0kg, a=2.0cm=0.2m`
(a) Frequency, `v=(1)/(T)=(1)/(2pi)sqrt((k)/(m))=(1)/(2xx3.14)sqrt((1200)/(3))=3.2s^(-1)`
(b) Acceleration, `A=omega^(2)y=(k)/(m)y`
Acceleration will be maximum when y is maximum, i.e., `y=a`
`:. ` Max. acceleration, `A_(max)=(ka)/(m)=(1200xx0.02)/(3)=8ms^(-2)`
(c) Max. speed ofo the mass will be when it is passing through the mean position, which is given by
`V_(max)=aomega=asqrt((k)/(m))=0.02xxsqrt((1200)/(3))=0.4ms^(-1)`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...