Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A 0.10 kg block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: x=(10 cm)cos[(10 rad​/​s) t+π​/​2 rad]. What is the maximum acceleration experienced by the block

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

10  m​/​s2

b

10 π m​/​s2

c

10π2 m​/​s2

d

10π3 m​/​s2

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

a=10×10−2m and ω=10 rad/secAmax=ω2a=10×10−2×102=10 m/sec2
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring
A 0.10 kg block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: x=(10 cm)cos[(10 rad​/​s) t+π​/​2 rad]. What is the maximum acceleration experienced by the block